Mahesh

Here students can locate TS Inter 2nd Year Physics Notes 3rd Lesson Wave Optics to prepare for their exam.

TS Inter 2nd Year Physics Notes 3rd Lesson Wave Optics

→ Huygens principle: Each point on the wave-front is the source of a secondary disturbance and the wavelets emanating from these points spread out in all directions with the speed of the wave. These wave lets emanating from the wavefront are usually referred to as secondary wavelets.
From Huygens principle every wave is a secondary wave to the preceding wave.

→ Wavefront: The locus of points which oscillate in phase is called “wavefront”.
(OR)
A wavefront is defined as a surface of constant phase.

→ Plane wave: Generally wavefront is spherical in nature when radius of sphere is very large. A small portion of spherical wave can be treated as plane wave.

→ Geometrical optics: It is a branch of optics in which we are completely neglects the finiteness of the wavelength is called geometrical optics. A ray of light is defined as the path of energy propagation. In this concept wavelength of light is tending to zero.

→ Snell’s law of refraction: Let nj and n2 are the refractive indices of the two media and i’ and ‘r’ are angle of incidence and angle of refraction then
n₁ sin i = n₂ sin r. This relation is called Snell’s law.
From Snell’s law for a given pair of media sin i / sin r = n₂ / n₁ is constant also called refractive index of the medium. Where nj is air or vacuum.

→ Critical angle (I_c): It is define as the angle of incidence in denser medium i for which angle of refraction in rarer medium r = 90°.
μ = \(\frac{1}{\sin c}\)
i. e, when r = 90° then i = C in denser medium.

→ Doppler’s effect in light: When there is relative motion between source and obser¬ver then there is a change in frequency of light received by the observer.

→ Red shift: If the source moves away from the, observer then frequency measured by observer is less (i.e., wavelength increases) as a result wavelength of received light moves towards red colour. This is known as “red shift”.

→ Blue shift: When source of light is approach¬ing the observer frequency of light received increases, (i.e., wavelength of light decreases.) As a result wavelength of received light will move towards blue colour. This is known as “blue shift”.

→ Superposition principle: According to super¬position principle at a particular point in the medium the resultant displacement produced by a number of waves is the vector sum of the displacements produced by each of the waves.

→ Coherent soures: Two sources are said to be coherent when the phase difference pro¬duced by each of two waves does not change with time.
Note: For a non – coherent waves the phase difference between them changes with time.

→ Interference: Interference is based on the superposition principle. According to which at a particular point in the medium the resultant displacement produced by a number of waves is the vecotor sum of displacements produced by each of the wave.
Note: In light when two coherent waves are superposed we will get dark and bright bands.

→ Constructive interference Or bright band:
When two coherent waves of path difference λ /2 or its integral multiples or a phase difference of or integral multiples of 2π are superposed on one another then the displacements of the two waves are in phase and intensity of light is 4I₀ where I₀ is intensity of each wave.
Condition, for constructive interference
is path difference = nλ or Φ = 0, 2π, 4π ………….. etc i.e., even multiples of π.

→ Destructive interference or dark band:
When two coherent waves of path difference \(\frac{\lambda}{2}\) or (n + \(\frac{1}{2}\)) λ or phase difference of \(\frac{\pi}{2}\) or odd multiples of \(\frac{\pi}{2}\) superposed at a given point then their displacements are out of phase and resultant intensity its is zero. This is called dark band.
For dark band to from path difference
= (n + \(\frac{1}{2}\)) λ (OR)
phase difference Φ = π, 3π, 5π …. odd multiples of n.

→ Diffraction: Bending of light rays at sharp edges (say edge of blade) is called”diffraction”.
As a result of diffraction we can see dark and bright bands close to the region of geometrical shadow.

→ Resolving power of Telescope:
Resolving power of telescope Δθ = \(\frac{0.61 \lambda}{\mathrm{a}}\)
Where 2a is aperture or vertical height (dia-meter) of lens used:
Resolving power of telescopes is its ability to show two distant object clearly when angular separation between them is Δθ.
When Δθ is less then resolving power of that telescope is high.
From above equation to increase resolving power of telescope its aperture or diameter of lens used must be high.

→ Resolving power of microscope:
The resolving power of the microscope is given by the reciprocal of the minimum separation of two points seen as distinct.
Minimum distance d_min = \(\frac{1.22 \lambda}{2 n \sin \beta}\), The term n sin β is called numerical aperture.
Resolving power of microscope = \(\frac{2 \mathrm{n} \sin \beta}{1.22 \lambda}\)
Note: The resolving power of microscope increases if refractive index n is high. In oil immersion objectives the lenses are placed in a transparent oil with refractive index close to that of objective lens to increase magnification.

→ Fresnel distance: The term z = a²/ λ is called fresnel distance.
In explaining the spreading of beam due to diffraction we will use the equation z = a²/ λ,
Where after travelling a distance z\(\frac{\lambda}{\mathrm{a}}\) size of beam is comparable to size of slit or hole ‘a’.
Note: When we are travelling from aperture to screen a distance z then width of diffrated beam z\(\frac{\lambda}{\mathrm{a}}\) is equals to aperture a’.
Beyond this distance ‘z’ divergence of the beam of width ‘a’ becomes significant. When distances are smaller than z spreading due to diffraction is small when compared with size of beam.

→ Polarisation: If is a process in which vib-rations of electric vectors of light are made ot oscillate, in a single direction.

→ Polaroids: A polaroid consists of a long chain of molecules aligned in a particular direction.

→ Malus’ Law: Let two polariods say P₁ and P₂ are arranged with some angle ‘θ’ between their axes. Then intensity of light coming
I = I₀ cos²θ
where I₀ is intensity of polarised light after passing through 1st polaroid P₁. This is known as Malus Law.

→ Uses of Polaroids: Polaroids are used

• to control intensity of light.
• They are used in photography,
• polaroids are used in sunglasses and in window panes.

→ Unpolarised light: In an unpolarised light electric vectors can vibrate in 360° direc¬tion perpendicular to direction of propaga¬tion. All these electric vectors can be grouped into two groups.

• Dot components they are vibrating perpendicular to the plane of the paper.
• Arrow components ‘↔’ i.e., their plane of vibration is along the plane of the paper.
  Thus an unploarised light can be shown as a group of dot components and arrow components.

→ Polorisation by scattering: The sky appears blue due to scattering of light. The light coming from clear blue portion of sky is made to pass through a polariser when it is rotated the intensity of light coming from polariser is found to be changing. Which shows that the scattered light consists of polarised light.

→ Polarisation by reflection:

• When unpolarised light falls on the boundary layer sepa-rating two transparent media the reflected light is found to be partially polarised. The amount of polarisation depends on angle of incidence i.
• It is found that when reflected ray and refracted ray are perpendicular the reflected ray is found to be totally plane polarised. The angle of incidence at this stage is known as Brewster angle.

→ Brewster’s angle: When reflected ray and refracted ray are mutually perpendicular then reflected ray is plane polarised. This particular angle of incidence i_B for which the reflected ray is plane polarised is called Brewster angle.

Explanation:
At Brewster angle i_B + r = \(\frac{\sin \mathrm{i}_{\mathrm{B}}}{\sin \mathrm{r}}=\frac{\sin \mathrm{i}_{\mathrm{B}}}{\sin \left(\pi / 2-\mathrm{i}_{\mathrm{B}}\right)}\)
n (OR) μ = \(\frac{\sin \mathrm{i}_{\mathrm{B}}}{\sin \mathrm{r}}=\frac{\sin \mathrm{i}_{\mathrm{B}}}{\sin \left(\pi / 2-\mathrm{i}_{\mathrm{B}}\right)}\)
= \(\frac{\sin \mathrm{i}_{\mathrm{B}}}{\cos \mathrm{i}_{\mathrm{B}}}\) = tani_B (OR) μ = tan i_B
∴ The tangent of Brewster’s angle tan (i_g) is equals to refractive index, i.e., μ = tan i_B.
Note: Refractive index can be shown by the symbol μ or n.

→ From the super position principle the resultant displacement is y = y₁ + y₂.
For constructive interference (bright band) y = y₁ + y₂) ; Intensity I = (y₁ + y₂)²
For destructive interference (dark band) y = y₁ ~ y₂; Intensity I = (y₁ – y₂)²

→ In interference, the resultant intensity
I = 4I₀ cos²\(\frac{\phi}{2}\) (Where I₀ is maximum intensity)
Resultant phase θ = \(\frac{\phi}{2}\) (Where Φ is initial phase difference)

→ Condition for formation of bright band is
(a) Path difference x = mλ, where m = 0,1, 2 ………….. etc.
(b) Phase difference Φ = 0, 2π ………… even multiples of π.

→ Condition for formation of dark band
(a) Path difference x = \(\frac{λ}{2}\) and odd multiples
(b) Phase difference Φ = π, 3π, 5π, ………….. odd multiples of n.

→ Relation between path difference (x) and
phase difference (Φ) is λ = \(\frac{\lambda}{2 \pi}\). Φ

→ Fringe with β = \(\frac{\lambda L}{\mathrm{~d}}\); Angular fringe width \(\frac{\beta}{L}=\frac{\lambda}{d}\)

→ Distance of mth bright band from central bright band is x₂ = \(\frac{\mathrm{m} \lambda \mathrm{L}}{\mathrm{d}}\)

→ Distance of m dark band from central dark band x₂ = \(\left(m+\frac{1}{2}\right) \lambda \frac{L}{d}\)

→ For two waves with intensities I and 12 with phase 4 resultant intensity
I = I₁ + I₂ + 2\(\sqrt{I_1 I_2}\)cos Φ

→ When a glass plate of thickness (t) is introduced in the path of one light wave then interference fringes will shift. Thickness glass plate t = \(\frac{m \lambda}{(\mu-1)}\)
m = number of fringes shifted;
λ = wavelength of light used.

→ When unpolarized light of intensity I, passes through a polarizer Intensity of emergent light I = \(\frac{\mathrm{I}_0}{2}\)

→ When polarized light falls on a polarizer with an angle θ to the axis then Intensity of refracted light I = I₀cos²θ

→ If polarized light falls on 1st polarIzer with an angle θ, and angle between the axes of given two polarizers is θ then intensity of light coming out of 2nd polarlzer I = I₀cos²θ₁cos²θ₂

→ For polarizatIon through reflection wIth Brewster angle i_B then μ or n = tan i_B

→ In diffraction radIus of central bright region
r₀ = \(\frac{1.22 \lambda \mathrm{f}}{2 \mathrm{a}}=\frac{0.61 \lambda \mathrm{f}}{\mathrm{a}}\)

→ ResolvIng power of telescope Δθ = \(\frac{0.61 \lambda}{\mathrm{a}}\)
Where Δθ Is the minimum angular separation between two distant objects.

→ Resolving power of microscope
d_min = \(=\frac{1.22 \mathrm{f} \lambda}{\mathrm{D}}=\frac{1.22 \lambda}{2 \tan \beta}=\frac{1.22 \lambda}{2 \sin \beta}=\frac{1.22 \lambda}{2 \mathrm{n} \sin \beta}=\frac{1.22 \lambda}{2 \mathrm{~N} \cdot \mathrm{A}}\)
Where d_min is the minimum separation required between two points in object.
N.A is numerical aperture (n sin β)
β is the angle subtended by the object at object lens.

→ Fresnel distance z = \(\frac{\mathrm{a}^2}{\lambda}\) Where ‘a’ is size of hole or slit.

Here students can locate TS Inter 2nd Year Physics Notes 2nd Lesson Ray Optics and Optical Instruments to prepare for their exam.

TS Inter 2nd Year Physics Notes 2nd Lesson Ray Optics and Optical Instruments

→ In ray optics, light will travel from one point to another point along a straight line. The path is called a “ray of light”. A bundle of such rays is called “a beam of light”.

→ Laws of reflection:

• Angle of reflection is equals to angle of incidence (r = i).
• The incident ray, the reflected ray and normal to the reflecting surface lie in the same plane.

→ Spherical mirrors:

• Pole (P): The geometrical centre of spherical mirror is called ‘pole ‘P’.”
• Principal axis: The line joining the pole (P) and centre of curvature ‘C’ of a spherical mirror is known as “principal axis”. Principal focus: For mirrors, after reflection a parallel beam of light w.r.to principal axis will converge or appears to diverge from a point on principal axis. This point is called “principal focus” ‘F’.
• Focal length (f): Distance between principal focus and pole of mirror or centre of lens is called “focal length”.

In spherical mirrors:

• Focal length of concave mirror is positive.
• Focal length of convex mirror is ve1.
• Relation between radius of curvature of mirror R and focal length f is R = 2f

→ Cartesian sign convention:

• All distances must be measured from the pole of the mirror or the optical centre of lens.
• The distances measured along the direction of the incident ray are taken as positive and the distances measured against the direction of incident ray are taken as negative.
• Distances measured above the principal axis of mirror (or) lens are taken as positive. The distances measured below are taken as ve’.

→ Mirror equation = \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\)

→ Linear Magnification (m): The ratio of height of image (ht) to height of object (hj is called “linear magnification”. Linear magnification (m) = \(\frac{\mathbf{h}_{\mathrm{i}}}{\mathrm{h}_{\mathrm{o}}}=\frac{-\mathrm{v}}{\mathbf{u}}\)

→ Refraction: The bending nature of light rays at refracting surface while travelling from one medium to another medium.

→ Laws of refraction:

• The incident ray, the refracted ray and normal to the interface at the point of incidence all lie in the same plane.
• Snell’s law: The ratio of sine of angle of incidence to sine of angle of refraction is constant for a given pair of media. sin i / sin r = n₂₁.
  where n₂₁ = refractive index of 2nd medium w.r.to 1st medium.
• When light rays are travelling from rarer medium to denser medium they will bend towards the normal.
• When light rays are travelling from denser medium to rarer medium they will bend away from normal.

→ Total internal reflection: When light rays are travelling from denser medium to rarer medium for angle of incidence i > i_c light rays are notable to penetrate the boundary layer and come back into the same medium. This phenomena is known as “total internal reflection”.

→ Critical angle (ic): The angle of incidence in the denser medium for which angle of refraction in rarer medium is 90° is called “critical angle” of denser medium.

→ Applications: Due to total internal reflection:

• Formation of mirages on hot summer days on tar roads and in deserts.
• Sparkling of well cut diamonds.
• Prisms designed to bend light rays by 90° or by 180° make use of total internal reflection.
• Optical fibre provides loss less trans-mission overlong distances with total internal reflection.

→ Refraction through lenses:
Lens formula I = \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\)
Lens makers formula \(\frac{1}{f}\) = (n₂₁ – 1) \(\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)
For any curved spherical surface \(\frac{\mathrm{n}_2}{\mathrm{v}}-\frac{\mathrm{n}_1}{\mathrm{u}}=\frac{\mathrm{n}_2-\mathrm{n}_1}{\mathrm{R}}\)

→ Power of a lens (P): Power of a lens is defined as the tangent of the angle by which it converges or diverges a beam of light falling at unit distance from the optical centre.
i.e, tan δ = \(\frac{h}{f}\) when δ is small tan δ = δ
δ = \(\frac{h}{f}\) (or) P = \(\frac{1}{f}\)
Power P = j unit: dioptre. Here, f is in meters.

→ Lens combination: When lenses are in contact then their combined focal length
\(\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}+\frac{1}{f_3}\)+…………
Combined power P = P₁ + P₂ + ……………

→ Magnification of combined lens is
m = m₁ × m₂ × m₃ i.e., total magnification is the product of individual magnifications of lenses on that combination.

→ Refraction through prism: In prism path of light ray inside prism is parallel to prism base.
Angle of prism, A = r₁ + r₂.
Angle of deviation δ = i + e – A
where e is angle of emergence.
At minimum deviation position
r₁ = r₂ ⇒ i₁ = r₂
r = \(\frac{A}{2}\) and I = \(\frac{\left(\mathrm{A}+\mathrm{d}_{\mathrm{m}}\right)}{2}\)
Refractive index n₂₁ = \(\frac{\mathrm{n}_2}{\mathrm{n}_1}\)
= \(\frac{\sin \left(\mathrm{A}+\mathrm{d}_{\mathrm{m}}\right) / 2}{\sin \mathrm{A} / 2}\)
For small angled prism dm = (n₂₁ – 1) A.

→ Dispersion: The phenomenon of splitting of light into its constitute colours is known as “dispersion”.
Dispersion takes place due to change in refractive index of medium for different wave lengths.

Note:

• The bending of red component of white light is least due to its longest wavelength.
• Bending of violet component of white light is maximum due to its short wave length.
• In vacuum all colours will travel with same velocity and velocity in vacuum is independent of wavelength of light.

→ Rainbow: Rainbow is due to dispersion of white light through water drops.

→ Scattering: Change in the direction of light rays in an irregular manner is called”scatte-ring”.
Ex: Sun light gets scattered by atmospheric particles.
Note: Light of shorter wavelengths will scatter much more than light of longer wave-length. Due to this, reason blue light will scatter more than red light.

→ Rayleigh scattering: The amount of scattering is inversely proportional to fourth power of the wavelength.
Ex: Blue has shorter wavelength than red. Hence blue will scatter much more strongly. Note: Wavelength of violet is shorter than blue. So violet will scatter more than blue. But our eye is not sensitive to violet. So we will see the blue colour in sky.

→ Human eye: Human eye contains rods and cones. Rods will respond to intensity of light.

→ Accommodation of eye: Our eye consists of a crystalline lens. The curvature and focal length of this lens is controlled by ciliary muscles. When ciliary muscles are relaxed focal length of eye lens is high. So light rays from distant object are focused onto the retina while viewing the near by objects eye lens will become thick and focal length will become less.
The property of eye lens to change its focal length depending on the objects to be viewed is called “accommodation”.

→ Hypermyopia: It is one type of eye defect. Where light rays coming from distant object are converged at a point infront of retina. This defect is called “short sighted-ness or Hypermyopia”. This defect can be compensated by using a concave lens.

→ Hypermetropia: This is one type of eye defect. In this defect eye lens focusses the incoming light at a point behind the retina. This is called “far sightedness or Hyper-metropia”. This defect can be compensated by using a convex lens.

→ Astigmatism: In this type of eye defect cornea will have large curvature in vertical direction and less curvature in horizon¬tal direction. A person with this type of defect can not clearly see objects because carnia has non-uniform radius.
This defect can be reduced by using cylindrical lenses of desired radius of curvature.

→ Simple microscope: A simple convex lens can be used as a simple microscope. Given object is placed between centre of lens and its focus. A virtual image is made to form at near point.
Near point magnification m = (1 + D/f).
If final image is made to form at infinity (far point).
Far point magnification m = D / f.

→ Compound microscope: Magnification of microscope
M = m₀ . m_e = \(\frac{v}{u}\left(1+\frac{D}{f_e}\right)\)
Or
m ≈ \(\frac{L}{f_e} \frac{D}{f_e}\), m₀ = \(\frac{h^{\prime}}{h}=\frac{v}{u}\)

For near point me = (1 + D/f_e)
For far point m = D/f_e.

→ Telescopes: It consists of two convex len¬ses mounted coaxially. Telescope is used to see large objects which are very far away.
Object lens will form a point size real image of object in between the lenses. This first image will act as an object to eye lens. Generally eye lens will form final image at infinity. This is called “Normal adjustment”.

Length of telescope L = f
Manification m = \(\frac{\beta}{\alpha}=\frac{\mathrm{f}_0}{\mathrm{f}_{\mathrm{e}}}\)
Reflecting telescopes are also called “cassegrain telescopes”.
In Telescopes if final image is inverted it is called Astronomical telescope. If final image is erect with respect to object it is called terrestrial telescope.

→ Velocity of light ¡n vacuum c = \(\frac{1}{\sqrt{\mu_0 \epsilon_0}}\)
= 3 × 10⁸ m/s
where μ₀ = permeability and E₀= permittivity of vacuum or air (free space).
In a medium the velocity of light (c) = \(\frac{v_1}{v_2}=\frac{v \lambda_1}{v \lambda_2}=\frac{\lambda_1}{\lambda_2}\)
also c = υλ

→ Refractive index of a medium μ = c / V;
where y = velocity of light in medium
From Sneils law μ = sin i / sin r.

→ In refraction or reflection the frequency of light remains constant
₁μ₂ or n₂₁ = \(\frac{v_1}{v_2}=\frac{v \lambda_1}{v \lambda_2}=\frac{\lambda_1}{\lambda_2}\)
Wavelength In medium λ_med = \(\frac{\lambda_{\text {vacuum }}}{\mu(\text { or }) n}\)

→ In refraction n₁λ₁ = n₂λ₂ (In refraction frequency of light υ is constant)

→ Relative refractive index n_r = \(\frac{\mathrm{n}_1}{\mathrm{n}_2}\)
Ex: _wn_g = \(\frac{\mathbf{n}_{\text {glass }}}{\mathbf{n}_{\text {water }}}=\frac{\lambda_{\text {water }}}{\lambda_{\text {glass }}}\)

→ In prism

• r₁ + r₂ = A (angle of prism);
• i₁ + i₂ = A + D (angle of deviation) (or) D = (i₁ – i₂) – A
• n = sin \(\frac{\mathrm{A}+\delta}{2}\)/sin(A/2)where δ = angle of minimum deviation.
• At minimum deviation position i₁ = i₂ and r₁ = r₂.
  So r = A/2 and i = \(\frac{\mathrm{A}+\delta}{2}\)

→ For small angled prisms 6 = (n – 1) A.

→ Relation between critical angle ‘c’ and refractive index n is
n = \(\frac{1}{\sin c}\) or c = sin^-1\(\left[\frac{1}{n}\right]\)

→ Lens makers formula \(\frac{1}{f}\) = (n – 1) \(\left[\frac{1}{n}\right]\)

→ Relation between object distance ‘u’, image distance y and focal length f is = \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\)

→ In lens combination \(\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}\) when in contact \(\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}-\frac{d}{f_1 f_2}\) when lenses are separated by a distance ‘d.

→ Condition to eliminate spherical aberration by lens combination is separation
d = f₁ – f₂.

→ Conditions to eliminate chromatic aberration:

• In lens combination with separation
  d = \(\frac{\mathrm{f}_1+\mathrm{f}_2}{2}\)
• In achromatic doublet \(\frac{\omega_1}{\mathrm{f}_1}+\frac{\omega_2}{\mathrm{f}_2}\) = 0

→ In simple microscope

• Near point magnification m = \(\left(1+\frac{D}{f}\right)\)
• When image is at Infinity m = \(\frac{D}{f}\)
  where D is the least distance of distinct vision.

→ In compound microscope

• Total magnification m = m₀ × m_e or m = \(\frac{v_0}{u}\left(1+\frac{D}{f_e}\right)\)
• When final Image is at infinity m = \(\frac{L D}{f_0 f_e}\)
• Object lens will form a real image, so m₀ = \(\frac{\mathrm{v}_0}{\mathrm{u}_0}\)
• Eye lens will form a virtual image at near point; so m₀ = (1 + \([latex]\)[/latex])

→ In Telescopes
(a) magnification m = \(\frac{\alpha}{\beta}=\frac{\tan \alpha}{\tan \beta}\) or m = \(\frac{\mathrm{f}_0}{\mathrm{f}_{\mathrm{e}}}\)
At normal adjustment, magnification
m = \(-\frac{\mathrm{f}_0}{\mathrm{f}_e}=\frac{\text { focal length of object lens }}{\text { focal length of eye lens }}\)

(b) When final image is at near point
m = \(-\frac{\mathrm{f}_0}{\mathrm{f}_{\mathrm{e}}}\left(1+\frac{\mathrm{f}_{\mathrm{e}}}{\mathrm{D}}\right)\)

(c) Length of telescope L = f₀ + f_e

(d) In terrestrial telescope length of telescope L = f₀ + f_e + 4f.

→ Dispersive power of prism
w = \(\frac{d_v-d_R}{\frac{d_v+d_g}{2}}=\frac{d_v-d_g}{d}=\frac{\mu_v-\mu_g}{(n-1)}\)
where d_v = Deviation of violet colour;
d_g = Deviation of red colour
\(\frac{\mathrm{d}_{\mathrm{v}}+\mathrm{d}_{\mathrm{g}}}{2}\) = Average deviation

→ Angular dispersive power to = (n_v – n_R)A

→ In prisms the condition for no dispersion is A₁ (n_g – n_R)₁ + A₂ (n_g – n_R)₂ = 0

→ In prisms the condition for no deviation is A₁ (n₁ – 1) + A₂ (n₂ – 1) = 0

→ In lens combination power of combined lens is P = P₁ + P₂ + ……………………

Here students can locate TS Inter 2nd Year Physics Notes 1st Lesson Waves to prepare for their exam.

TS Inter 2nd Year Physics Notes 1st Lesson Waves

→ Wave: A wave is a physical manifestation of disturbance that propagates in space.

→ Transverse waves: In these waves, the con-stituents of the medium will oscillate perpen-dicular to the direction of propagation of the wave.

→ Longitudinal waves: In these waves the constituents of the medium will oscillate paral¬lel to the direction of propagation of the wave.

→ Wave motion can be represented as a func¬tion of both position ‘x’ and time ‘t’.

→ Generally for x – ‘+ve’ direction equation of a wave is y = a sin (kx – ωt – Φ)

→ Crest: It is a point of a maximum positive displacement.

→ Trough: It is a point of maximum negative displacement.

→ Amplitude (a): The maximum displacement of constituents of the medium from means position is called “amplitude” ‘a’.
a = y_max

→ Phase (Φ) Phase gives the displacement of the wave at any position and at any instant.

→ Initial Phase: At initial condition (when x = 0 and t = 0) phase of the wave is called initial phase.

→ Wavelength (λ): The minimum distance between any two successive points of same phase on wave is called “wavelength” A.

→ Propagation constant (OR) angular wave number (k):
\(\frac{2 \pi}{\lambda}\) or A = \(\frac{2 \pi}{\mathrm{k}}\) where ‘A’ is A k

→ Time period (T): Time taken to produce one complete wave (or) time taken to complete one oscillation is known as “Time period T”.

→ Frequency ‘υ’: Number of waves produced per second. (OR) Number of oscillations com¬pleted per second is known as “frequency υ.”
Frequency υ = \(\frac{1}{\mathrm{~T}}\) (Or) Time period T = \(\frac{1}{v}\)

→ Angular frequency (or) velocity (ω):
Angular frequency (w) = 2πυ (or) ω = \(\frac{2 \pi}{\mathrm{T}}\)

→ Relation between velocity v, wavelength A and frequency o is v = υλ (OR) v = λ/T.

→ Speed of a wave in stretched strings:
Speed of transverse wave in stretched wires v =\(\sqrt{\mathrm{T} / \mu}\)
where µ = linear density = mass / length. S.I. unit = kg/metre.

→ Speed of longitudinal waves in different media
In liquids:
v = \(\sqrt{\frac{B}{\rho}}\)
where B = Bulk modulus
ρ = Density.

→ In solids:
v = \(\sqrt{\frac{Y}{\rho}}\)
where Y = Young’s modulus
ρ = Density.

In gases:
According to Newton’s formula
v = \(\sqrt{\frac{P}{\rho}}\)
where P = pressure of the gas and
ρ = density of the gas.
According to Newton – Laplace’s formula
v = \(\sqrt{\frac{\gamma \mathrm{P}}{\rho}}\) where γ = the ratio of specific heats of the gas.

→ Principle of superposition: If two or more waves moving in the medium superposes then the resultant wave form is the sum of wave functions of individual waves.
i.e y = y₁ + y₂ + y₃ + ………

→ Stationary waves (or) standing waves :
When a progressive wave and reflected wave superpose with suitable phase a steady wave pattern is set up on the string or in the medium.
A standing wave is represented by y (x, t) = 2 a sin kx cos ωt.

→ Fundamental mode : The lowest possible natural frequency of a system is called “fundamental mode.”

→ Frequency of fundamental mode is called fundamental frequency (or) first harmonic.

→ Harmonics: Sounds with frequencies equal to integral multiple of a fundamental frequency (n) are called “harmonics.”

→ Resonance : It is a special condition of a system where frequency of external periodic force is equal to or almost equal to natural frequency of a vibrating body.

→ Beats: When two sounds of nearly equal frequencies are produced together they will pro-duce a waxing and waning intensity of sound at observer. This effect is called Heats.”
Beat Frequency Δυ = υ₁ ~ υ₂
Beat Period T = \(\frac{1}{v_1 \sim v_2}\)

→ Doppler’s effect: The apparent change in frequency of sound heard due to relative motion of source and observer is called” Doppler’s effect.”

→ Doppler’s effect is applicable to mechanical waves and also to electromagnetic waves. In sound it is “asymmetric” whereas in light it is “symmetric”.

→ Velocity of sound in a medium v = υλ
where υ = \(\frac{1}{T}\)

→ Propagation constant of wave (k) = \(\frac{2 \pi}{\lambda}\) (Also k is known as angular wave number) Angular velocity of wave (©) = \(\frac{2 \pi}{\lambda}\) = 2πυ; frequency υ = \(\frac{1}{T}=\frac{2 \pi}{\omega}\)

→ Equation of progressive wave in x-positive direction is
y’= a sin (ωt – kx) (or) y = a cos (ωt – kx)
Along – ve direction on X-axis y = a sin (ωt + kx) (or) y = a cos (ωt + kx)

→ From the superposition principle, the dis-placement of the resultant wave is given by y = y₁ + y₂

→ Equation of stationary wave is
y = 2 A sin kx cos ωt or Y = 2A kx sin ωt Here kx and ωt are in separate trigonometric functions.

→ In stretched wires of string
(i) Velocity of transverse vibrations
v = \(\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mu}}\)
where T = tension applied
and ρ = linear density,

(ii) Fundamental frequency of vibration
υ₀ = \(\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mu}}\)

→ The laws of transverse vibrations in stretched strings

• 1st law, υ ∝ \(\) (OR) \(\frac{v_1}{v_2}=\frac{l_2}{l_1}\)
• 2nd law υ ∝ √T (OR) \(\frac{v_1}{v_2}=\sqrt{\frac{\mathrm{T}_1}{\mathrm{~T}_2}}\)
• 3rd law υ ∝ \(\frac{1}{\sqrt{\mu}}\) (OR) \(\frac{v_1}{v_2}=\sqrt{\frac{\mu_2}{\mu_1}}\)

→ Newton’s equations for velocity of sound in different media.
1. In solids υ_s = \(\sqrt{\frac{Y}{\rho}}\)
Y = Young’s modulus of wire

2. In liquids υ₁ = \(\sqrt{\frac{B}{\rho}}\)
B = Bulk modulus of liquid

3. In gases υ_g = \(\sqrt{\frac{\mathrm{P}}{\rho}}\)
P = Pressure of the gas Laplace corrected the formula for velocity of sound in gases as υ_g = \(\sqrt{\frac{\gamma \mathrm{P}}{\rho}}\)
where γ = \(\frac{C_P}{C_V}\)
γ = Ratio of specific heats of a gas.
Where Y = Young’s modulus of solid,
K = Bulk modulus of the liquid and P is pressure of the gas.

→ In case of closed pipes

• Length of pipe at the fundamental note is l = \(\frac{\lambda}{4}\) ⇒ λ = 4l
• Fundamental frequency of vibration
  υ = \(\frac{\mathrm{v}}{\lambda}=\frac{\mathrm{v}}{4 l}\), υ’ = 3υ, υ” = 5u.
  υ’ and υ” are second harmonic and third harmonics.
• Closes pipes will support only odd har-monics.
  Ratio of frequencies or harmonics is 1: 3: 5: 7 etc.

→ In case of open pipes .
1. Length of pipe at the fundamental note is l,
l = \(\frac{\lambda}{2}\) ⇒ λ = 2l

2. Fundamental frequency of vibration
υ = \(\frac{\mathrm{v}}{\lambda}=\frac{\mathrm{v}}{2 l}\)
υ’ = 2υ, υ” = 3υ.
υ’ and υ” are second and third harmonics.

3. Open pipe will support all harmonics of a fundamental frequency.
Ratio of frequencies = 1: 2: 3: 4

→ Beat frequency Δυ = υ₁ ~ υ₂

→ When a tuning fork is loaded, its frequency of vibration decreases.
Due to loading, beat frequency decreases ⇒ frequency of that fork υ₁, < υ₂ (2nd fork).
When a tuning fork is field then its frequency of vibrating increases.
Due to filing beat frequency increases ⇒ frequency of that fork υ₁ > υ₂ (2nd fork).

→ General equation for Doppler’s effect is
υ’ = \(\left[\frac{v \pm v_0}{v \mp v_s}\right]\)υ
When velocity of medium (v^ is also taken into account apparent frequency
υ’ = \(\left[\frac{v+v_0 \pm v_m}{v \mp v_s \mp v m}\right]\)υ Sign convention is to be applied.
Note: In sign convension direction from observer to source is taken as + ve direction of velocity.

Students must practice these Maths 2A Important Questions TS Inter Second Year Maths 2A Complex Numbers Important Questions Very Short Answer Type to help strengthen their preparations for exams.

TS Inter Second Year Maths 2A Complex Numbers Important Questions Very Short Answer Type

Question 1.
Find the additive inverse of (√3, 5). [March ’90]
Solution:
Let z = (√3, 5)
The additive inverse of z is – z = – (√3, 5) = (- √3, – 5)

Question 2.
If z = (cos θ, sin θ), find (z – \(\frac{1}{z}\)). [TS – Mar. 2019]
Solution:
Given that z = (cos θ, sin θ)
The multiplicative inverse of z is
z^-1 = \(\frac{1}{z}=\left(\frac{a}{a^2+b^2}, \frac{-b}{a^2+b^2}\right)\)
= \(\left(\frac{\cos \theta}{\cos ^2 \theta+\sin ^2 \theta}, \frac{-\sin \theta}{\cos ^2 \theta+\sin ^2 \theta}\right)\)
= \(\left(\frac{\cos \theta}{1}, \frac{-\sin \theta}{1}\right)\)
= (cos θ – sin θ)
z – \(\frac{1}{z}\) = (cos θ, sin θ) – (cos θ, – sin θ)
= (cos θ – cos θ, sin θ + sin θ)
= (0, 2 sin θ)

Question 3.
Find the multiplicative inverse of (sin θ, cos θ).
Solution:
Let z = (sin θ, cos θ)
The multiplicative inverse of z is
z^-1 = \(\left(\frac{a}{a^2+b^2}, \frac{-b}{a^2+b^2}\right)\)
z^-1 = \(\left(\frac{\sin \theta}{\cos ^2 \theta+\sin ^2 \theta}, \frac{-\cos \theta}{\cos ^2 \theta+\sin ^2 \theta}\right)\)
= \(\left(\frac{\sin \theta}{1}, \frac{-\cos \theta}{1}\right)\)
= (sin θ, – cos θ)

Question 4.
Express \(\frac{4+2 i}{1-2 i}+\frac{3+4 i}{2+3 i}\) in the form a + ib.
Solution:

Question 5.
Find the real and imaginary parts of the complex number \(\frac{a+i b}{a-i b}\). [TS – Mar. 2015]
Solution:

Question 6.
Express (1 – i)³ (1 + i) in the form a + ib.
Solution:
Let z = (1 – i)³ (1 + i)
= (1 – i)² (1 – i) (1 + i)
= (1 + i² – 2i) (1 – i²)
= (1 – 1 – 2i) (1 – (- 1)) (∵ i² = – 1)
= (- 2i) (2)
= – 4i = 0 + i (- 4)
It is in the form a + ib.

Question 7.
Find the multiplicative inverse of 7 + 24i. [TS – Mar. 2016, AP – May 2015]
Solution:
Let z = 7 + 24i
z^-1 = \(\)
∴ The multiplicative inverse of complex number z = 7 + 24i is
z = \(\left(\frac{7-24 i}{7^2+(24)^2}\right)\)
= \(\left(\frac{7-24 \mathrm{i}}{49+576}\right)=\frac{7-24 \mathrm{i}}{625}\) .

Question 8.
If 4x + i(3x – y) = 3 – 6i where x and y are real numbers, then find the values of\ x and y.
Solution:
Given that 4x + i(3x – y) = 3 – 6i
Comparing real and imaginary parts on both sides
4x = 3
x = \(\frac{3}{4}\)

3x – y = – 6
3 (\(\frac{3}{4}\)) – y = – 6
y = \(\frac{9}{4}\) + 6
= \(\frac{9+24}{4}\)
= \(\frac{33}{4}\)
∴ x = \(\frac{3}{4}\), y = \(\frac{33}{4}\)

Question 9.
Find the complex conjugate of (3 + 4i) (2 – 3i). [May ’14, Mar. 1997, AP – Mar. ’18, May ’16]
Solution:
Let z = (3 + 4i) (2 – 3i)
= 6 – 9i + 8i – 12i²
= 6 – i + 12
z = 18 – i
∴ The conjugate of z is \(\overline{\mathrm{z}}=\overline{(18-\mathrm{i})}\) = 18 + ¡

Question 10.
Find the square root of – 5 + 12i. [Board Paper]
Solution:

Question 11.
Find the square root of – 47 + i. 8√3. [March ’13(old), May ’12, ’09, May ’10, ’03]
Solution:

Question 12.
Simplify i² + i⁴ + i⁶ + …………… + (2n + 1) terms.
Solution:
= i² + i⁴ + i⁶ + …………… + (2n + 1) terms
= i² + (i²)² + (i²)³ + …………….. + (2n + 1) terms
= – 1 + 1 – 1 + ……………. + (2n + 1) terms = – 1

Question 13.
Simplify i¹⁸ – 3 . i⁷ + i² (1 + i⁴) (- i)²⁶.
Solution:
i¹⁸ – 3 . i⁷ + i² (1 + i⁴) (- i)²⁶.
= (i²)⁹ – 3 (i²)³ . i + i² (1 + (i²)²) (i²)¹³
= (- 1)⁹ – 3 (- 1)³ . i + (- 1) (1 + (- 1)²) (- 1)¹³
= – 1 + 3i + 2 = 1 + 3i

Question 14.
If (a + ib)² = x + iy, find x² + y². [TS – May 2015]
Solution:
Given that,
(a + ib)² = x + iy
a² + i²b² + 2abi = x + iy
a² – b² + i(2ab) =x + iy
Comparing real and imaginary parts on both sides
x = a² – b², y = 2ab
∴ x² + y² = (a² + b²)² + 4a²b² = (a² + b²)²

Question 15.
Find the least positive integers n, satisfying \(\left(\frac{1+i}{1-i}\right)^n\) = 1.
Solution:

If n = 4
⇒ i⁴ = 1
∴ The required least positive integer, n = 4.

Question 16.
Write z = – √7 + i √21 in the polar form. [AP – May 2015; March 11]
Solution:

Let, z = – √7 + i √21
= r (cos θ + i sin θ)
then r cos θ = – √7, r sin θ = √21
r = \(\sqrt{x^2+y^2}=\sqrt{(-\sqrt{7})^2+(\sqrt{21})^2}\)
= \(\sqrt{7+21}=\sqrt{28}=2 \sqrt{7}\)
Hence,
2√7 cos θ = – √7
cos θ = \(-\frac{1}{2}\)
2√7 sin θ = √21
2√7 sin θ = √3 . √7
sin θ = \(\frac{\sqrt{3}}{2}\)
∴ θ lies in the Q2.
∴ θ = \(\frac{2 \pi}{3}\)
∴ z = – √7 + i√21
= 2√7 (cos \(\frac{2 \pi}{3}\) + i sin \(\frac{2 \pi}{3}\)).

Question 17.
Express – 1 – i in polar form with principle value of the amplitude.
Solution:

Let, – 1 – i = r (cos θ + i sin θ)
Then r cos θ = – 1; r sin θ = – 1
r = \(\sqrt{(-1)^2+(-1)^2}=\sqrt{1+1}=\sqrt{2}\)
Hence,
√2 cos θ = – 1
cos θ = \(\frac{-1}{\sqrt{2}}\)
∴ θ lies in the Q3.
∴ θ = – \(\frac{3 \pi}{4}\)

√2 sin θ = – 1
sin θ = \(\frac{-1}{\sqrt{2}}\)

∴ – 1 – i = √2 (cos (- \(\frac{3 \pi}{4}\)) + i sin (- \(\frac{3 \pi}{4}\)))

Question 18.
Express – 1 – i√3 in the modulus amplitude form.
Solution:

Let – 1 – i√3 = r (cos θ + i sin θ)
Then r cos θ = – 1, r sin θ = – √3
r = \(\sqrt{\mathrm{x}^2+\mathrm{y}^2}\)
= \(\sqrt{(-1)^2+(-\sqrt{3})^2}\)
= \(\sqrt{1+3}=\sqrt{4}\) = 2
Hence, 2 cos θ = – 1, 2 sin θ = – √3
cos θ = – \(\frac{1}{2}\), sin θ = \(\frac{-\sqrt{3}}{2}\)
∴ θ lies in the Q3.
∴ θ = – \(\frac{2 \pi}{3}\)
∴ For the given complex number modulus = 2
Principle amplitude = – \(\frac{2 \pi}{3}\)
∴ – 1 – i √3 = 2 [cos (- \(\frac{2 \pi}{3}\)) + i sin (- \(\frac{2 \pi}{3}\))]

Question 19.
If z₁ = – 1 and z₂ = – 1 then find Arg(z₁z₂).
Solution:
Given, z₁ = – 1
= cos π + i sin π
∴ Arg z₁ = π
z₂ = – i
= cos (- \(\frac{\pi}{2}\)) + i sin (- \(\frac{\pi}{2}\))
∴ Arg z₂ = – \(\frac{\pi}{2}\)
Now,
Arg (z₁z₂) = Arg z₁ + Arg z₂
= π – \(\frac{\pi}{2}\) = \(\frac{\pi}{2}\)

Question 20.
If z₁ = – 1, z₂ = i, then find Arg (\(\frac{z_1}{z_2}\)).
[May 14 11, March 09 Board Paper TS-Mar. 16; AP – Mar. 18, 17]
Solution:
Given z₁ = – 1
= cos π + i sin π
∴ Arg z₁ = π
z₂ = i
= cos \(\frac{\pi}{2}\) + i sin \(\frac{\pi}{2}\)
∴ Arg z₂ = \(\frac{\pi}{2}\)
Now, Arg \(\left(\frac{\mathrm{z}_1}{\mathrm{z}_2}\right)\) = Arg z₁ – Arg z₂
= π – \(\frac{\pi}{2}\) = \(\frac{\pi}{2}\)

Question 21.
If (cos 2α + i sin 2α) (cos 2β + i sin 2β) = cos θ + i sin θ, then find the value of θ.
Solution:
Given,
(cos 2α + i sin 2α) (cos 2β + i sin 2β) = cos θ + i sin θ
cis (2α) . cis (2β) = cis θ
cis (2α + 2β) = cis θ
cos (2α + 2β) + i sin (2α + 2β) = cos θ + i sin θ
Comparing real parts on both sides, we get
cos θ = (2α + 2β)
θ = 2 (α + β)

Question 22.
If √3 + i = r (cos θ + i sin θ) then find the value of θ in radian measure.
Solution:

Given that,
√3 + i = r (cos θ + i sin θ)
Then r cos θ = √3, r sin θ = 1
r = \(\sqrt{x^2+y^2}=\sqrt{(\sqrt{3})^2+(1)^2}\)
= \(\sqrt{3+1}=\sqrt{4}\) = 2
Hence,
2 cos θ = √3
cos θ = \(\frac{\sqrt{3}}{2}\)
2 sin θ = 1
sin θ = \(\frac{1}{2}\)
∴ θ lies in the Q1.
∴ θ = \(\frac{\pi}{6}\)

Question 23.
If x + iy = cis α . cis β, then find the value of x² + y². [AP – Mar. 18]
Sol.
Given that,
x + iy = cis α . cis β
= cis (α + β)
= cos (α + β) + i sin (α + β)
Now, comparing real parts on both sides.
we get x = cos (α + β)
Comparing imaginary parts on both sides
we get y = sin (α + β)
Now,
x² + y² = cos² (α + β) + sin² (α + β) = 1.

Question 24.
If (√3 + i)¹⁰⁰ = 2⁹⁹ (a + ib) then show that a² + b² = 4. [AP – Mar.2016] [AP – Mar. 2019]
Solution:
Given that,
(√3 + i)¹⁰⁰ = 2⁹⁹ (a + ib)

Squaring on both sides,
a² + b² = 4

Question 25.
If z = x + iy and |z| = 1, then find the locus of z. [TS – Mar. 2019]
Solution:
Given, z = x + iy
|z| = 1
|x + iy| = 1
\(\sqrt{x^2+y^2}\) = 1
Squaring on both sides,
x² + y² = 1
∴ Locus of z is x² + y² = 1.

Question 26.
If the amplitude of (z – 1) is \(\frac{\pi}{2}\) then find the locus of z. [TS – May 2015]
Solution:
Let, z = x + iy
Now, z – 1 = x + iy – 1
= (x – 1) + iy
If z = x + iy then θ = tan^-1 (\(\frac{y}{x}\))
Given that,
the amplitude of (z – 1) is \(\frac{\pi}{2}\)
tan^-1 (\(\frac{y}{x-1}\)) = \(\frac{\pi}{2}\)
\(\frac{y}{x-1}\) = tan \(\frac{\pi}{2}\)
\(\frac{y}{x-1}\) = ∞ = \(\frac{1}{0}\)
x – 1 = 0
x = 1
∴ Locus of z is x = 1.

Question 27.
If the Arg \(\overline{\mathbf{z}}_1\) and Arg z₂ are \(\frac{\pi}{5}\) and \(\frac{\pi}{3}\) respectively, then find (Arg z₁ + Arg z₂). [AP-May, Mar. 2016]
Solution:
If z = x + iy then θ = tan^-1 (\(\frac{y}{x}\))
If \(\overline{\mathrm{Z}}\) = x – iy then θ = tan^-1 (\(\frac{-y}{x}\))
= – tan^-1 (\(\frac{y}{x}\))
Given that,
Arg \(\overline{\mathbf{z}}_1\) = \(\frac{\pi}{5}\)
Arg z₁ = – \(\frac{\pi}{5}\)
Arg z₂ = \(\frac{\pi}{3}\)
Now,
Arg z₁ + Arg z₂ = – \(\frac{\pi}{5}\) + \(\frac{\pi}{3}\)
= \(\frac{-3 \pi+5 \pi}{15}=\frac{2 \pi}{15}\).

Question 28.
If z = \(\frac{1+2 i}{1-(1-i)^2}\) then find Arg(z).
Solution:
z = \(\frac{1+2 i}{1-(1-i)^2}\)
= \(\frac{1+2 \mathrm{i}}{1-\left(1+\mathrm{i}^2-2 \mathrm{i}\right)}\)
= \(\frac{1+2 i}{1-1-i^2+2 i}\)
= \(\frac{1+2 \mathrm{i}}{1+2 \mathrm{i}}\) = 1
∴ z = 1 = 1 + i(0)
∴ Arg z = tan^-1 (\(\frac{y}{x}\))
= tan^-1 (\(\frac{0}{1}\))
= tan (0) = 0°

Question 29.
Simplify \(\frac{(2+4 i)(-1+2 i)}{(-1-i)(3-i)}\) and find its modulus.
Solution:
Given,

Question 30.
Simplify \(\frac{(1+i)^3}{(2+i)(1+2 i)}\) and find its modulus.
Solution:
Given, \(\frac{(1+i)^3}{(2+i)(1+2 i)}\)

Question 31.
If (1 – i) (2 – i) (3 – i) ……….. (1 – ni) = x – iy, then prove that 2 . 5 . 10 …………….. (1 + n²) = x² + y².
Solution:
Given that,
(1 – i) (2 – i) (3 – i) (1 – ni) = x – iy

Squaring on bothsides,
2 . 5 . 10 . …………… (1 + n²) = x² + y².

Question 32.
If the real part of \(\frac{z+1}{z+i}\) is 1, then find the locus of z.
Solution:
Let, z = x + iy
Now,

Given that,
the real part of \(\frac{z+1}{z+i}\) is 1.
⇒ \(\frac{x^2+x+y^2+y}{x^2+(y+1)^2}\) = 0
x² + y² + x + y = x² + y² + 2y + 1
x – y – 1 = 0
∴ The locus of z is x – y – 1 = 0.

Question 33.
If |z – 3 + i| = 4 determine the locus of z. [March 14, May 08]
Solution:
Let z = x + iy
Given, |z – 3 + i| = 4
|x + iy – 3 + i| = 4
|(x – 3) + i (y + 1)| = 4
\(\sqrt{(\mathrm{x}-3)^2+(\mathrm{y}+1)^2}\) = 4
Squaring on both sides
(x – 3)² + (y + 1)² = 16
x² + 9 – 6x + y² + 2y + 1 – 16 = 0
x² + y² – 6x + 2y – 6 = 0
∴ The locus of z is x² + y² – 6x – 2y – 6 = 0.

Question 34.
If |z + ai| = |z – ai|, then find the locus of z.
Solution:
Let, z = x + iy
Given, |z + ai| = |z – ai|
|x + iy + ai| = |x – iy – ai|
|x + i(y + a)| = |x + i(y – a)|
\(\sqrt{(x)^2+(y+a)^2}=\sqrt{(x)^2+(y-a)^2}\)
Squaring on both sides
x² + (y + a)² = x² + (y – a)²
(y + a)² – (y – a)² = 0
4ya = 0
y = 0
∴ The locus of z is y = 0.

Question 35.
Find the equation of the straight line joining the points represented by (- 4 + 3i), (2 – 3i) in the Argand plane.
Solution:
Let the two complex numbers be represented in the Argand plane by the points P, Q respectively.
Then P = (- 4, 3), Q = (2, – 3)
∴ The equation of \(\overline{\mathrm{PQ}}\) is
y – y₁ = \(\frac{\mathrm{y}_2-\mathrm{y}_1}{\mathrm{x}_2-\mathrm{x}_1}\) (x – x₁)
y – 3 = \(\frac{-3-3}{2+4}\) (x + 4)
y – 3 = \(\frac{-6}{6}\) (x + 4)
y – 3 = – 1 (x + 4)
y – 3 = – x – 4
x + y + 1 = 0.

Question 36.
z = x + iy represents a point in the Argand plane. Find the locus of z such that z = 2.
Solution:
Given, z = x + iy
=> P = (x, y)
|z| = 2
|x + iy| = 2
\(\sqrt{\mathrm{x}^2+\mathrm{y}^2}\) = 2
Squaring on both sides
x² + y² = 4
x² + y² – 4 = 0
∴ Locus of P is x² + y² – 4 = 0
The locus represents a circle with centre (0, 0) and radius 2 units.

Question 37.
The point ‘P’ represents a complex number z in the Argand plane. If the amplitude of z is \(\frac{\pi}{4}\), determine the locus of P.
Solution:
Let z = x + iy
⇒ P = (x, y)
Given that,
Amp(z) = \(\frac{\pi}{4}\)
tan^-1 (\(\frac{y}{x}\)) = \(\frac{\pi}{4}\)
\(\frac{y}{x}\) = tan \(\frac{\pi}{4}\)
\(\frac{y}{x}\) = 1
y = x
x = y
∴ Locus of P is x = y i.e., x – y = 0.
The locus represents a line passing through origin and making an angle of 45° with the (+) ve direction of X – axis.

Question 38.
Find the equation of the perpendicular bisector of the line segment joining the points 7 + 7i, 7 – 7i in the Argand plane.
Solution:

Let the two complex numbers be represented in the Argand plane by the points P, Q respectively.
Then P = (7, 7), Q = (7, – 7)
Since, C is the midpoint of \(\overline{\mathrm{PQ}}\) then
C = \(\left(\frac{\mathrm{x}_1+\mathrm{x}_2}{2}, \frac{\mathrm{y}_1+\mathrm{y}_2}{2}\right)\)
= \(\left(\frac{7+7}{2}, \frac{7-7}{2}\right)=\left(\frac{14}{2}, \frac{0}{2}\right)\) = (7, 0)
Slope of \(\overline{\mathrm{PQ}}\) is m = \(\frac{y_2-y_1}{x_2-x_1}\)
= \(\frac{-7-7}{7-7}=\frac{-14}{0}\)
∵ \(\overline{\mathrm{AB}}\) is ⊥ to \(\overline{\mathrm{PQ}}\) then
slope of \(\overline{\mathrm{AB}}\) = \(\frac{-1}{\mathrm{~m}}=\frac{-1}{\frac{-14}{0}}\) = o
The equation of the straight line \(\overline{\mathrm{AB}}\) is
y – y₁ = \(-\frac{1}{m}\) (x – x₁)
y – 0 = 0 (x – 7)
⇒ y = 0.

Question 39.
Find the equation of the straight line joining the points – 9 + 6i, 11 – 4i in the Argand plane.
Solution:
Let the two complex numbers be represented in the Argand plane by the point P, Q respectively.
Then P = (- 9, 6), Q = (11, – 4)
∴ The equation of \(\overline{\mathrm{PQ}}\) is
y – y₁ = \(\frac{y_2-y_1}{x_2-x_1}\) (x – x₁)
y – 6 = \(\frac{-4-6}{11+9}\) (x + 9)
y-6 = \(\frac{-10}{20}\) (x + 9)
y – 6 = \(\frac{-1}{2}\) (x + 9)
2y – 12 = – x – 9
x + 2y – 3 = 0.

Question 40.
Show that the points in the Argand diagram represented by the complex numbers 2 + 2i – 2 – 2i, – 2√3 + 2√3i are the vertices of an equilateral triangle. [AP – May 15, 07, TS-Mar. 18]
Solution:
Let the three complex numbers be represented in the Argand plane by the points P, Q, R respectively.
Then P = (2, 2), Q = (- 2, – 2). R = (- 2√3, 2√3)
Now,
PQ = \(\sqrt{(2+2)^2+(2+2)^2}\)
= \(\sqrt{(4)^2+(4)^2}=\sqrt{16+16}=\sqrt{32}\)

QR = \(\sqrt{(-2+2 \sqrt{3})^2+(-2-2 \sqrt{3})^2}\)
= \(\sqrt{4+12-8 \sqrt{3}+4+12+8 \sqrt{3}}=\sqrt{32}\)

PR = \(\sqrt{(2+2 \sqrt{3})^2+(2-2 \sqrt{3})^2}\)
= \(\sqrt{4+12+8 \sqrt{3}+4+12-8 \sqrt{3}}=\sqrt{32}\)

∴ PQ = QR = PR
∴ Given points form an equilateral triangle.

Question 41.
If z = 2 – 3i, then show that z² – 4z + 13 = 0. [TS- Mar. 18, Mar. 08, AP – Mar. 2019]
Solution:
Given that z = 2 – 3i
z – 2 = – 3i
Squaring on both sides
(z – 2)² = (- 3i)²
z² + 4 – 4z = 9i²
z² – 4z + 4 = – 9
∴ z² – 4z + 13 = 0

Question 42.
Find the multiplicative inverse of (3, 4).
Solution:
(\(\frac{3}{25}\), \(\frac{4}{25}\))

Question 43.
Write \(\frac{2+5 i}{3-2 i}+\frac{2-5 i}{3+2 i}\) in the form of a + ib.
Solution:
\(\frac{-8}{13}\) + i (0)

Question 44.
Write the complex number (1 + 2i)³ in the form a + ib. [TS – Mar. 2017]
Solution:
– 11 – 2i

Question 45.
Find the multiplicative inverse of √5 + 3i.
Solution:
\(\frac{\sqrt{5}}{14}-\frac{3}{14} i\)

Question 46.
Write the conjugate of (2 + 5i) (- 4 + 6i).
Solution:
– 38 + 8i.

Question 47.
Find the square roots of 3 + 4i. [March ’13 (old). May ’12, ’09, May ’10, ’03]
Solution:
± (2 + i)

Question 48.
Find the square roots of 7 + 24i. [TS – May 2016, Mar. ‘14]
Solution:
± (4 + 3i)

Question 49.
Find the square roots of – 8 – 6i.
Solution:
± (1 – 3i)

Telangana TSBIE TS Inter 2nd Year Political Science Study Material 1st Lesson Indian Constitution-Historical Background Textbook Questions and Answers.

TS Inter 2nd Year Political Science Study Material 1st Lesson Indian Constitution-Historical Background

Long Answer Questions

Question 1.
Explain the causes for the Origin of Indian National Movement.
Answer:
Indian National Movement was organized during 1857 -1947 in both Violent and Non – Violent forms. The First war of Independence in 1857 known as a Great Revolt by Indian Soldiers sowed the seeds of Nationalism among Indians. With the formation of Indian National Congress in 1885, the movement for National Independence took a Non – Violent form. The National Movement united people of different groups to fight against the British Repression.

The Following are the causes for the birth of Indian National Movement:
1) British Colonial Rule :
British colonial rule is said to be one of the important cause for the birth and growth of nationalism in India. The British rulers developed communication facilities such as road, rail, press, posts and telegraphs for preserving their administrative, political and economic interests in India. The leaders of Indian National Movement properly availed such facilities and informed the Indian masses about tyrannic and exploitative policy of the British colonial rulers. The political unity rendered by the British rulers also brought the feelings of unity among the Indians.

2) Socio – Cultural Renaissance :
Several Social and Cultural Movements were witnessed in India during the 19^th century which brought social awakening and the ideas of Cultural Nationalism. The Brahma Samaj founded by Raja Rammohan Roy, pioneered these movements and gave a call for reform against social evils such as Sati, Child marriages, Widowhood, Idol Worship, Seclusion and others. This was followed by Arya Samaj, The Ramakrishna mission, The Theosophical Society, Prartana Samaj, the Aligarh movement, Satyasodhak Samaj, and Wahabi Movement. These movements have infused the ideas of rationalism, socio-cultural identity and patriotism and indirectly motivated the people to have an urge for self-rule.

3) Great Revolt:
1857 Revolt was the first great challenge to the British rule in India. The Revolt inspired the struggle for Indian Independence. It gave courage to Indians against the British.

4) English Education :
English was introduced in India as a medium of instruction and correspondence by the British rulers on the recommendations of ford Macaulay committee. A few educated elite among Indians travelled in England to gain proficiency in English language. They were able to study and understand the views of eminent thinkers like Hobbes, Locke, Rousseaue, J.S.Mill and Karl Marx on several concepts like liberty, equality/fraternity, independene, democracy etc. They communicated the value and importance of these ideas to Indians in various regional languages This inspired the Indian masses to participate in the freedom struggle.

5) Economic Exploitation :
The British government transformed India into a safe citadel for British investments, markets and export of Indian raw materials. It suppressed the growth of Indian cottage, village and small scale industries for safe guarding the British investments on permanent basis. The British Governor generals like Lord Lytton and Curzon, implemented such policies suited to the British investors exploiting the national wealth of India. This has caused bitter resentment among the Indian industrialists, cottage and small scale artisans. Hence, they were attracted towards the freedom movement in India.

6) Famines and Acute poverty:
During 19^th century several famines took place, but the British government had not done anything for the people. Poverty is another cause. Number of people died because of starvation. This led to great dissatisfaction among the Indians and enabled them to fight against the British government.

7) Press :
Many Journals and dailies were published and circulated in India during freedom movement. Ex : Amrit Bazar, Kesari, Marata, Hindu, Andhra Patrika etc. They created, enhanced national feelings among themselves. They played a crucial role in spreading National feelings and conveyed the desires and demands of people to the Britishers.

8) Repressive Rule :
The Arms Act and Vernacular Press Act passed by Lord Lytton and his Kabul invasion hurt the sentiments of Indians. The llbert Bill Proposed by Lord Rippon and its withdrawl made the Indians understand the racial hatred of the British.

9) Racial Discrimination :
Indians were treated as second rate citizens and were excluded from higher posts. The British ploicy of racial discrimination aroused and strengthened Nationalism in India.

10) Emergence of the Indian National Congress :
1885 was a landmark in the history of Indian National Movement. A.O. Hume, the British Civil Servant established the Indian National Congress on 28^th December, 1885 at Mumbai in Gokul Tejpaul Sanskrit College and developed the nationalist feelings among the people. W.C. Banerjee was the first President of the Congress.

Question 2.
Describe the various phases of Indian National Movement.
Answer:
Stages of the Indian National Movement : Dr. Ramesh Chandra an eminent historian, has divided the history of Indian National Movement into three stages, namely :

1. Moderate stage (1885 – 1905)
2. Extremist stage (1905 – 1920)
3. Gandhian stage (1920 – 1947)

Let us examine these three stages in detail.
1) Moderate stage :
The early leaders of the Congress are known as the “Moderates”. They dominated the first stage of the freedom movement from 1885 to 1905. They had full faith in the British sense of justice and fair play. They emphasised the use of peaceful and constitutional methods to achieve their aims and objectives. They did not believe in agitation or unconstitutional methods. They carried on their work by means of public debates, propaganda, petitions, demonstrations and deputations.Their motto was “reform, not revolution”.The prominent leaders of this group are W.C.Benerjee, Pherozshah Mehta, Dadabhai Baoroji, Surendranath Benerjee, Bodaruddin Toyabji, Dinshaw Wacha R.E. Dutt., L.M. Chose, Ranade, G.K. Gtokhale etc.

Demands of Moderates : The demands of the Moderates are :

1. Reduction in the military expenditure.
2. Abolition of the Indian Council.
3. Holding Civil Service Examinations in India simultaneously with that of England.
4. Expansion of the Legislative councils so as to include more and more Indians.
5. Separation of judiciary from executive.
6. Reduction in land revenue and granting occupancy rights to the tillers.
7. Irrigation facilities to the farmers.
8. Stoppage of export of food grains.
9. Higher jobs to the Indians.
10. Reduction in taxes.

2. Extremist stage (1906 -1919):
The second stage of the National Movement was dominated by the extremists from 1906 to 1918. During this period the congress entered the militant stage. The prominent leaders of this group are Bala Gangadhar Tilak, Lala Lajpat Roy and Bipin Chandrapal. They advocated the use of militant methods.They raised the slogans of “Swaraj” and “Swadesi” and Laid stress on National Education.

Demands of Extremists :
Extremists regarded the British rule as a curse for India. They had faith in the superiority of Vedic culture, literature language and civilisation. They demanded the introduction of self government in the country. They did not demand petty concessions but favoured complete Swaraj. They gave the people the slogan that “Freedom was their birth right and they must have it”.

The moderates preferred action to peaceful methods for achieving the goals. They had no faith in petitions and representations. They laid emphasis on boycott of foreign goods and use of Swadesi goods.

3. Gandhian stage (1920 -1947):
The third stage of the National Movement was dominated by Mahatma Gandhi from 1920 to 1947. That is why this period is known as the “Gandhian period”. He launched his first non-violent and non-cooperation movement in 1920.

The movement included the boycott of foreign goods, legislative councils, law Courts, schools and colleges. But, when the movement look a violent course, he suspended it abruptly in 1922. In 1924, he vvas elected as congress president and called upon to organise demonstrations against Simon Commission in 1927. In 1929, the Congress at its Lahore session declared its goal as achievement of’complete independence”.

In March, 1930 Gandhiji started his second agitation known as “Civil disobedience Movement” or “Salt Satyagraha”. Gandhiji asked the British to “Quit India” and gave a call of “Do or Die”.

After the Second World -War, the labour party, headed by Clement Atlee came to power in England. He sent cabinet mission to India. On the basis of cabinet mission plan and Mountbatten plan, the British) parliament passed the Indian Independence Act. India became independent on 15th August, 1947.

Question 3.
Critically analyse the provisions in the Government of India Act 1935.
Answer:
The Government of India Act, 1935 was considered to be the second mile stone in the introduction of the responsible government in India. Different factors like Indian Nationalism, British imperialism, communal tendencies etc, influenced in a way or the other in adopting this Act. Similarly various factors, like reports of Muddiman committee (1924), Simon Commission (1927), the deliberations of three Round Table Conferences (1930,1931 & 1932), Communal Award (1932), White Paper (1933), Report of Joint Parliamentary Committee (1934) etc., formed bases for adopting this Act. This Act was a detailed and lengthy one, containing 14 parts and 10 schedules.

Main Provisions:

1. The Act prescribed Quasi – rigid and Quasi – flexible methods to amend the provisions.
2. While it abolished dyarchy in the provinces, it introduced the same at the union level.
3. It provided provincial autonomy in the states and paved the way for a complete responsible rule.
4. It split the administrative subjects of the Union Government into two lists – Reserved and Transferred. While the reserved list comprised Defence, External affairs, Tribal welfare, Communal issues, the remaining subjects were brought under transferred category.
5. The Act has enlarged the sphere of electorate and various legislatures.
6. It provided for bicameralism at the centre and in six out of eleven provinces.
7. It constituted a federal court in order to settle the conflicts between the union and the provinces or between the provinces.
8. It proposed an All India Federation, having two governments (union and states), division of powers, written and rigid constitution, federal court etc. It included 59 items in federal or central list and 54 items in provincial list. It has incorporated certain other subjects like civil and criminal procedure code, Marriage, Divorce, Endowments, Contracts, Press etc, in the concurrent list.
9. It gave freedom to the native provinces either to join or Quit the Federation.
10. The Act has brought about far reaching changes in the home administration in India. It abolished the Indian Council and authorized the secretary of state to appoint 3 to 6 members to assist in his duties.
11.  New provinces Orissa and Sindh were created.
12. Separate Representation for Indian Christians, Anglo Indians, Europeans and Depressed classes in Legislative Councils.

Criticism :
This Act Could not satisfy the Nationalist aspirations of the people for both political and economic power continued to be concentrated in the hands of the British. This Act was criticised for providing despotic powers to governor general. Indian National Congress Condemned this Act and demanded for Complete Independence and formation of a constituent Assembly to draft a New Constitution.

Question 4.
Explain the salient features of Indian Constitution.
Answer:
Introduction:
The Indian Constitution was prepared and Adopted by the Constituent Assembly, which was set up in 1946. The Constituent Assembly took nearly three years From 9^th December,1946 to 25^th November, 1949 (2 years, 11 months and 18 days) to complete the framing of the Costitution. The Constituent Assembly approved the Indian Constitution on 26^th November, 1949. The Indian Costitution came into force on 26 January, 1950, which we have been celebrating as “The Republic Day”. The following are the salient or basic features of the Indian Constitution.

1) Written and Detailed Constitution :
The Constitution of India is a written document. It consists of 12 Schedules, 22 Parts and 444 Articles. Many factors have contributed for the bulkiness of the Constitution. All most all matters relating to the composition and organization of union, states as well as Union Territories, provisions regarding protection of interests of Schedules castes, Scheduled Tribes and other Backward classes, provisions regarding Special Constitutional bodies like the Election Commissions, the UPSC and State Public Service Commissions are made in a very comprehensive. manner. Similarly, several matters relating to Fundamental Rights, Fundamental Duties, Directive principles of state policy, Union – State Relations, Official Language and Regional Languages were clearly mentioned in the Costitution.

2) India is a Sovereign, Socialist, Secular, Democratic, Republic:
The constitution deliberately designed India to be a Sovereign, Socialist, Secular, Democratic, Republic. Each of these concepts has wide ranging dimensions.

Sovereign State :
India is a Sovereign State. It is clear that India is internally Supreme and externally independent of any foreign control. India is not a sub-ordinate State.

Socialist State :
According to the preamble India is committed to Socialism assuring the establishment of a socialist State. It strives to provide social and economic Justice to all its people and end all forms of exploitation. We follow mixed economy. The State is free to bring the key industries and the private enterprises under the state ownership and management.

Secular State :
Secularism means that the State protects all religions equally and does not itself subscribe to any other religion as official.

Democratic State :
Democracy is the formidable foundation of Indian political system. It states that the people of India are the chief source of the Political Authority they can make and unmake the government. The people are both the Rulers and the ruled. India adopted Indirect or Representative Democracy.

Republican State :
India is a Republic with an elected Head of the State i.e., the President of India. No public office in the country is hereditary. Any Indian citizen can contest to any office of the government subject to certain limitations.

3) Noble Aims and objectives :
Indian constitution has noble aims and objectives. The Indian constitution secures to all its citizens, Justice – Social, Economic, and Political matters. It also provides all its citizens, liberty of thoight, expression, belief, faith, and worship. Further, the constitution of India promotes among all people the equality of status and opportunity. Like wise it strives for fraternity by assuring the dignity of the individual and the unity and integrity of the Nation.

4) Combination of ligidity and Flexibility :
The Constitution of India can be amended with Rigid and Flexible method. Article 368 provides the details of the amendment procedure. It can be amended by three methods.

Firstly:
The Unio l Parliament can amend some parts of the constitution by a simple majority. Ex: The formation of new ates (Ex: Telangana), changing the boundaries of states, provisions relating to citizenship etc. It is said o be Flexible.

Secondly :
Some provisions can be amended by a special majority i.e., not less than Two Thirds (2/3) of the member of the House present and voting.
Ex : Fundamental Rights, Directive principles of state policy, etc. It is said to be half rigid and half flexible.

Thirdly:
Some provisions can be amended by Two – Thirds (2/3) majority of the parliament and with the concurrence c half of the states.
Ex: Election of the president, Executive powers of the union and the states, Distribution of Legislative powers between the union nd the states ete. lt is said to be Rigid.

Hence, the Constitu1 on has a combination of Rigid and Flexible methods to amend the Constitution.

5) Unitary and Federal Features :
India is a union of states according to the constitution. Our constitution contains both th features of unitary and Federal Governments. It prescribed unitary system in emergencies and federal lystem on ordinary occussions. Provisions of unitary state such as Single Citizenship, Single IntegratedBudiciary, Single election Commission, Roll of All India Services personnel etc., are found in our constitut on. At the same time certain federal features like written, Rigid constitution, Dual Government, Bicamenlism etc., are profoundly seen in our constitution. Thus it is a quasi – federal policy like Canada.

6) Parliamentary government:
The makers of our constitution adopted the Irish precedent of of a hereditary Monarch. They retained all other essential features of two types of executive heads, Prime Ministers leadership, collective’ responsibility, nominal position of the President, Parliaments control over the Union Executive etc. They prescribed the same type of political system for the states.

7) Independent Judiciary :
The constitution of India granted independent and integrated judiciary for Indians. Hence t ie Supreme Court and High Courts in India act independently without subject to the control of the executive and legislative organs, judiciary enjoys independence in the matters of appointment of judges, their tenure, salaries and allowances, service conditions, promotion etc. Our constitution clearly states that the executive and legislative authorities should refrain from interfering in the functioning of the judiciary. It conferred the power of judicial review on the judges of the Supreme Court and High Courts. It is due to the independent position that the Supreme Court and High Court judges will decide the propriety and constitutional validity of the acts and policies of the legislative and executive authorities in the country.

8) Directive Principles of State Policy:
Our constitution hinted our certain directive principles as the policy of the state in Part IV from Articles 36 to 51. The makers of our constitution derived these principles from Irish constitution. They pointed out that these principles would transform India into a welfare, gandhian and liberal oriented state. These principles must be implemented by all the parties which hold authority without political considerations.

These principles enable the state to provide a new social order based on economic, political and social justice. These principles include certain programmes like provision of employment opportunities, fair distribution of wealth, equal pay for equal work, educational and child care for thbse children below fourteen years, etc. Similarly unemployment relief, old age pension, protection against ill health, provsion of leisure for workers, conservation of wild life etc are included in these principle.

9) Fundamental Rights :
Our constitution mentioned fundamental rights on American model for the complete realization of the personality of Indian citizens. They are incorporated under Part III from Articles 12 to 35 of our constitution. Indian citizens can utilize these rights subject to certain rational restraints. Nobody including the government is allowed to interfere in these rights. The higher judicial organizations in the country help the citizens in safeguarding these rights. At first there were seven fundamental rights in the constitution. But at present there are only six fundamental rights. They are 1. Right to Equality 2. Right to Freedom 3. Right against Exploitation 4 Right to Religion 5. Cultural and Educational Rights and 6. Right to Constitutional Remedies.

10) Fundamental Duties :
Our constitution incorporated fundamental duties in Article 51 A under Part 4A. At first there were 10 fundamental duties inserted through the Constitution (Forty second) Amendment Act, 1976. Later one more duty was added through the Constitution (Eighty sixth) Amendment Act, 2002. Altogether there are Eleven Fundament Duties in our consititution. Respecting the consititution, National Flag, National Anthem, Safeguarding public property etc are come of the Fundamental duties.

11. Single Citizenship :
Our Constitution provides for single citizenship for all persons who are born in India and who resided in India for a specific period. It enables the citizens to possess and enjoy identical right and privileges. It also promotes unity, integrity and fraterrity among the people.

12. Universal Adult Franchise:
The makers of the Indian Constitution provided for the Universal Adult Franchise for all citizens without any discrimination based on caste, colour, creed, community, language, religion, region, sex, property etc. At the beginning, Adult Franchise was given to all the citizens who attained the age of 21 years. Later voting age was reduced to 18 years through the 61^st Constitution Amendment Act in 1988.

13. Bicameralism :
The Constitution of India introduced Bi-cameralism at the National level. Accordingly, the Indian Parliament consists of two houses namely the Rajya Sabha (upper house) and the Lok Sabha (Lower house). While the Rajya Sabha represents the states, the Lok Sabha represents the people.

14. Panchayati Raj and Nagar Palika Acts :
The Panchayati Raj and Nagar Palikas Acts are recent features of our constitution. The 73^rd and 74^th constitutional recognition to the rural and urban local governments which came into force in 1993 and 1994 respectively had become operative all over the territory of India. The ideas of democratic decentralisation or the grassroof democracy are realised by these acts. These Acts provides for adequate representation for Womem, Scheduled Castes, Scheduled tribes and other Weaker Sections in the policy making bodies of the Local governments.

15. Special Provisions Relating to Scheduled Castes and Scheduled Tribes :
Indian Constitution hinted out certain specific directives for the development of scheduled castes and scheduled tribes in India. It enabled the union and state governments to review the steps taken for the improvement of backward classes through setting up o^ independent commissions. In this regard the union government was authorised to act as a coordinator between various state governments.

Short Answer Questions

Question 1.
Explain any four causes for the Indian National Movement.
Answer:
The British rule contributed to the emergence of Modern Indian Nationalism. The National movement united people of different groups to fight the Britishers. The following are the some causes for the birth of Indian National Movement.

1. British Colonial Rule :
British colonial rule is said to be one of the important cause for the birth and growth of Nationalism in India. The British rulers developed communication facilities such as road, rail, press, posts and telegraphs for preserving their administrative, political and economic interests in India. The leaders of Indian National Movement properly availed such facilities and informed the Indian masses about the tyrannic and exploitative policy of the British colonial rulers. The political unity rendered by tbe British rulers also brought the feelings of unity among the Indians.

2. English Education :
English was introduced in India as a medium of instruction and correspondence by the British rulers on the recommendations of ford Macaulay committee. A few educated elite among Indians travelled in England to gain proficiency in English language. They were able to study and understand the views of eminent thinkers like Hobbes, Locke, Rousseaue, J.S. Mill and Karl Marx on several concepts like liberty, equality, fraternity, independence, democracy etc.

3. Economic Exploitation :
The British government transformed India into a safe citadel for British investments, markets and export of Indian raw materials. It suppressed the growth of Indian cottage, village and small scale industries for safe guarding the British investments on permanent basis. The British Governor generals like Lord Lytton and Curzon, implemented such policies suited to the British investors exploiting the national wealth of India.This has caused bitter resentment among the Indian industrialists, cottage and small scale artisans. Hence, they were attracted towards the freedom movement in India.

4. Press :
Many Journals and dailies were published and circulated in India during freedom movement. Ex : Amrit Bazar, Kesari, Marata, Hindu, Andhra Patrika etc. They created, enhanced national feelings among themselves. They played a crucial role in spreading, National feelings and conveyed the desires and demands of people to the Britishers.

Question 2.
Explain the role of the extremists in Indian National Movement.
Answer:
The second stage of the National Movement (from 1906 to 1919) was dominated by the extremists. During this period, the Congress entered the militant stage. The prominent leaders were Lala Lajpat Roy, Bala Gangadhar Tilak and Bipin Chandra pal (Lai – Bal – Pal). The extremists criticised the moderates for pursuing the policy of “Mendicancy with a begging bowl” and advocated-the use of radical methods to realise Swaraj.

The extremists proposed new methods to achieve their objective! of self government. They declared openly that they had no faith in the goodness of the British. The extremists advocated (1) Boycott of foreign goods (2) Swadesi (3) Boycott of the Government controlled schools and colleges (4) Promoting National education and (5) Passive resistance.

The extremists appealed to the people not to assist the Government to rule. They wanted to create trouble to the Government to rule. They wanted to create trouble to the Government in collecting revenue. They were also against fighting beyond the frontiers or outside India with Indian blood and money. They even advocated that people must organise parallel courts to die British courts. Boycott of foreign goods and the use of Swadesi goods were popularised by the extremists.

They also adovacated boycott of British courts, municipalities and legislative councils. The extremists popularised education on national lines. In national education, the focus was on things Indian. The regional Indian language was to be the chief medium of instruction. The message of Swadesi began to reah the masses due to the methods used by the Extremists.

The extremists introduced new methods of political organisation and new ways of continuing political struggles. But the methods of passive resistance and non-co-operation remained mere ideas. It must also be said that over emphasis on Hindus religious symbols during the extremist National Movement created a rift between the Hindus and the Muslims. The mass of the common people were also outside the mainstream of national politics during the extremist movement despite their radical methods.

Question 3.
Write about the important events during the Gandhian phase of Indian National Movement.
Answer:
Gandhian phase (1920-1947) is the last and final stage of Indian National Movement. Gandhiji led many movements againts the British such as the Non Co-operation movement, Civil disobedience movement and ‘Quit India movement. They may be explained in the following ways.

1. Non Co-operation Movement (1920 to 1922) :
Gandhiji launched this Non-cooperation movement on August 1920 as a protest against the incidents of kilafat and Jallian Wallabagh. The Indian National Congress presided over by Nagapur session in 1921 headed by Lala Lajapathi Roy. There relate to a) Positive programmes like collecting Tilak fund Rs. one Crore and distributed 20 lakhs of charakas. b) Nagative programmes like renouncing British titles, honarary offices, Boycotting courts, schools, legislative councils, as a part Gandhi renounced his title Kaiker-E-Hind. But unfortunately this non-co-operation movement called off by Gandhi in 1922 due to the bad incident happened at Chouri-Choura. ‘

2. Civil Disobedience Movement (1930 to 1934):
The Indian National Congress lanunched this movement on 13^th March, 1930 under the guidence of Gandhi. During the 1^st stage Gandhiji asked the Indian masses to launch Salt Satyagraha. Gandhiji with his 78 staunch disciples launched a march from Sabarmathi to Dandi nearly 240 miles on 13^th March 1930. In the second stage people voluntarily organised hartals, packetings, boycott, swadesi and other programmes. In the third stage between Aug. 1933 to May 1934 in this stage leaders launched collective and individual civil disobedience. Finally Gandhiji concentrated on non-political activites.

3. Quit India movement (1942 to 1944):
The Congress Working Committee at its meeting in worda on July 14, 1942 demanded the British Government on Quit India. Again the Congress in its meeting in Bombay 8^th August 1942 approved the Quit India movement. In its 1^st stage Britishers arrest Gandhi then National wide processions, demonistrations taken agitators destroyed police stations, railway stations and post offices. Peaceful conditions prevailed during the last stage. Quit India movement continued from 8^th August, 1942 with the imprisonment of Gandhi to May 1944 with his release. Through this movement the British government too undestood. The patriotic feelings of the people. They conceded to grant Independence to Indians.

Question 4.
Describe the Home Rule Movement in India’s Freedom Struggle.
Answer:
Home Rule Movement is an important movement organized during the Indian freedom struggle. Lokmanya Bala Gangadhar Tilak and Annie Besant led this movement. The British government freed Tilak from Mondalay prison after six years of imprisonment. Tilak who returned to India decided to renew harmonious relations with Indian National Congress leaders and to behave positively with the British rulers through some constructive activities. He strived to form home rule councils on the pattern of Irish Councils.

He maintained contacts with the Irish Woman, Annie Besant of Madras (now Chennai), the founder of Divine Life Society. In 1916 both of them separately set up Home Rule Leagues and tried to create political awareness and spiritual enlightment among the people. Annie Besant founded New India and Common Wheel and appointed George Arundale as the organizational secretary of Home Rule League.

Tilak propagated the ideals of Home Rule League in Bombay (excepting Maharashtra), Karnataka, Central Province and Bihar. On the other hand Annie Besant spread the above ideals in the remaining parts of India. The two leaders propagated secular ideas preaching anti-caste and antireligious elements among the people. They extensively distributed home rule league pamphlets, throughout the country. The British government imprisoned Tilak in July, 1916 on the pretext that his conduct was not good. This enraged the followers of Tilak. Mohammad Ali Jinnah acted as the pleader for Tilak in district and High Courts and tried to bring him out of prison. Tilak intensified his agitation. While he founded home rule legue offices in six places, Annie Besant set up such offices in more than 200 places.

Question 5.
What are the various programmes adopted during the Non-cooperation Movement?
Answer:
Gandhiji started Non-cooperation Movement on 1^st August, 1920. The Nagpur session of the Congress in 1920 ratified it. The programme of the Movement was as follows.

1. Boycott of foreign goods and use of Swadeshi goods.
2. Surrender of titles and honorary offices and resignation from nominated seats in local bodies.
3. Boycott of Government and State-aided schools and colleges and establishment of National schools and colleges.
4. Boycott of elections to the new Councils and refusal by the voters to vote at the elections.
5. Refusal to attend official Durbars.
6. Gradual boycott of British courts by lawyers and citizens.
7. Refusal by soldiers, clerks and working people to serve in Mesopotamia.

The Non-cooperation Movement, in short, aimed at the boycott of colleges, courts, councils and Government jobs. Besides the boycott measures, the Congress also adopted a constructive programme. It included the opening of national educational institutions, the setting up of panchayats as a substitute of British courts, the popularisation of chakra palying etc.

Question 6.
What are the important provisions of Government of India Act,1919?
Answer:
The Govt, of India Act, 1919 or the Montague Chelmsford reforms made a beginning in the representative Govt, of India. Discontent over 1909 reforms, frustration among Indians, changes in the Muslim league policy, the first world war, the reunion of moderates and extremists in the congress, the Home rule movement, Montague’s declaration, Montford scheme etc., were some of the causes led to the adoption of this Act.

Main Provisions:

1. The Act removed the restrictions over the size of membership of Governor General’s Executive Council.

2. The Act made the executive responsible to the Central Legislative Assembly.

3. It fixed the membership of the council of states at 60 and that of the Central Legislative Assembly at 145 respectively.

4. It has provided considerable powers to the Central Legislative Assembly by empowering the latter to make laws on union list subject to the approval the Governor-General.

5. It has provided Dyarchy at the state level and divided provincial subjects into reserved subjects and transferred subjects. 28 subjects were included in the reserved category. Finance, Land Revenue, Justice, Police, Irrigation, Factories, Industries etc., were a few among them. The Governors of the provinces were to administer these subjects with the help of the British councillors.

The Act has listed 22 subjects in the transferred category. Local self Govt Agriculture, Public works, Public health, Co-operation, Education etc., were some such subjects. The Governors were to administer these subjects with the help of the Indian Ministers.

6. The Act has created die office of the secretary of State which acts as the agent of the British Govt in India.

7. The Act further divided the administrative subjects between the union and state governments into two types namely: (i) Union list and (ii). Provincial list It included 47 items in the union list Defence, Foreign affairs, Public debt, Posts and Telegraphs, Navigation, Export and Import Duties etc., were a few of such subjects. It has incorporated 51 items in the provincial list. Local government, Public health, Education, Agriculture, Forests, Law and order etc., formed part of this list.

8. The Act separated provincial budgets from that of the union. It empowered the provincial legislatures to approve their own budgets from that of the union. It empowered the provincial legislatures to approve their own budgets and levy taxes.

9. The Act introduced some changes in the composition of Indian Council. The size of Indian council was fixed between 8 and 12 its tenure at five years.

10. The Chamber of princes with viceroy as the chairman was constituted. The Govt of India Act, 1919 was considered to be a landmark in the constitutional history of India.

Question 7.
Point out the main provisions of the Independence of India Act, 1947.
Answer:
The main provisions of the Independence of the India Act 1947 may be explained as follows.

1. India and Pakistan shall be constituted as two separate independent states.
2. Two constituent Assemblies shall be formed each for India and Pakistan and work as constitution making bodies as well as legislative bodies in their respective domains.
3. The British Supermacy over British India native princely states shall cease to exit. The princely states would be free to join either dominious or to remain independent.
4. Territories of the two dominous were defined but they were empowered to include or exlude a territory themselves.

Question 8.
Write about any three basic features of Indian Constitution.
Answer:
1. Single citizenship :
Our Constitution provides for single citizenship for all persons who are bom in India and who resided in India for a Specific period. It enables the citizens to possess an enjoy identical rights and privilages. It also promotes unity, integrity and fraternity among the people.

2. Universal adult Franchise :
The makers of the Indian constitution provided for the universal adult franchise for all citizens without any discrimination based on caste, colour, creed, community, language, religion, region, sex, property etc., At the beginning, adult franchise was given to all the citizens who attained the age of 21 years. Later voting age was reduced to 18 years through the Constitution Amendment Act in 1988.

3. Bicameralism :
The constitution of India introduced Bi – cameralism at the national level. Accordingly, the Indian Parliament consists of two houses namely the Rajya Sabha (upper house) and Lok Sabha (lower house). While the Rajya Sabha represents the states, the Lok Sabha represents the people.

Very Short Answer Quenstions

Question 1.
Moderates in Indian National Movement.
Answer:
Dadabhai Naoroji, Copala Krishna Gokhale, Surendra Nath Benarjee, Umesh Chandra Benarjee are the prominent moderate leaders. They adopted peaceful and constitutional methods in articulating their demands in the National Movement. There include prayers, petitions protest and mediatious.

Question 2.
Methods of Extremists.
Answer:

1. Boycott of British goods, honorary titles and Govt., offices.
2. Encouraging native education.
3. Striving for membership to Indian in Legislative Councils.
4. Patronizing native goods and industries.
5. Practicing passive resistance.

Question 3.
Simon Commission.
Answer:
The Commission had seven English men as members and Sir John Simon as Chairman. Hence, the Congress decided to boycott the Commission’s prceedings throughout the country and raised the slogan “Simon Go back”. Indians wholeheartedly obliged the call and protested against thes Simon Commission. Black flag demonstrations greated the Commission wherever it went. In Lahore, the boycott demonstrations were led by lala Lajpat Rai who became a victim of police latticharge that resulted in his death three weeks later.

Question 4.
Civil Disobedience Movement.
Answer:
The Civil Disobedience Movement of 1930-31 is sometimes called the Salt Satyagraha. Mahatma Gandhi, compelled by circumstances, wrote a letter to Lord Irwin, the then Viceroy of India stating that he would launch a Civil Disobedience Movement by breaking salt laws, if the Government did not concede his demands. As there was no favourable response from the Viceroy, Gandhi started his famous Dandi March. Accompanied by 78 workers he marched on foot from Sabarmati Ashram to the seashore on 12th March, 1930. On 6th April,1930, he started the Civil Disobedience Movement by picking salt laying on the seashore. The programme of the movement included

1. The Violation of Salt law.
2. Abstention from attending educational institutions by students and the offices by the public servants.
3. Picketing of shops dealing in liquor,opium and foreign goods.
4. Bonafire of foreign cloth.
5. Non-payment of taxes.

The movement gathered momentum very soon.Thousands of people participated in it. The Civil Disobedience Movement aimed at paralysing the administrative by performance of specific illegal acts.

Question 5.
Minto – Morley Reforms Act.
Answer:
Lord Minto the very of India, Lord Morley the Secretary of State for India played a key role in the formulation of this Act. This Act passed the way by introducing significant changes in the structure and maintainance of legislative bodies in India. The British’s Government considered several factors while formulating this Act.

Question 6.
Constituent Assembly.
Answer:
There are 309 members in the constituent assembly of them 296 members belongs to British India and 93 belongs to princely states. Elections for constituent assembly held on July – August -1946. Dr. Babu Rajendra prasad elected as the chairman for constituent assembly. Drafting Committee was formed on 29th August, 1947. Dr. Ambedhkar was appointed as the chairman.

Question 7.
Drafting Committee.
Answer:
A seven member Drafting Committee was constituted on August 29, by the Constituent Assembly Prominent members like Sir Alladi Krishna Swamy Ayyar, N. Gopala Swami lyangar, Sayyed Mohammad Sadullah, Dr.K.M. Munshi, B.L. Mittar and D.R Khaithan were the six members in this committee. B.N. Rao was the advisor and S.N. Mukherjee, the chief draftsman. The drafting committee has submitted the draft constitution to the Constituent Assembly on November 5,1947. The draft was published on February 21, 1948.

Question 8.
Rigid and flexible features of Indian Constitution.
Answer:
Some provisions like election of the President, Powers of Union and State Governments, Supreme Court and High Courts require rigid amendment procedure. Some other provisions like changes in the names, boundaries, areas and separation of states etc., may be amended by a simple majority in Parliament, some provisions like Fundamental Rights and Directive Principles of State policy are amended by a half rigid and half flexible method.

Question 9.
Preamble of the Indian constitution.
Answer:
The Indian Constitution begins with a preamble. The preamble clearly defines the objectives of our constitution. It declares India as a sovereign, socialist, secular democratic republic it provides Liberty, Equality Fraternity and Justice. It states that the people of India are the cheif sources of the Political Authority.

Question 10.
Parliamentary Government.
Answer:
India follows Parliamentary form at the Central and the State levels. In this form of government nominal executive (President or Governor) functions on the advice of the real executive (RM. or C.M heading the Council of Ministers). The other features of this government are : Collective responsibility, membership of the ministers in the legislature, no fixed term of office for the cabinet, leadership of the RM. etc.

Question 11.
Unitary and Federal features of Indian Constitution.
Answer:
Unitary Features :

1. Single Citizenship
2. Integrated Judiciary
3. Appointment of Governors
4. Role of All India Services personnel in state administration etc., are unitary features.

Federal Features :

1. Written and Rigid constitution.
2. Dual Government
3. Bi-cameralism etc are Federal Features of the Indian Constitution.

Question 12.
Universal Adult Franchise.
Answer:
it means granting of the right to vote to all adult citizens who have completed a particular age. In India all citizens who have completed 18 years of age have this right to vote. Generally, lunatics, criminals, insolvents are excluded from this right. It is the backbone of Democracy.