Resolving power of telescope is defined as the reciprocal of the smallest angular separation between two distant objects whose images are to be seen separately.

When the prism is held in water 
\({ ^{ w }\mu }_{ g }=Sin\left( A+m/2 \right) /SinA/2\)
As \(^{ w }{ \mu }_{ R }< { ^{ a }\mu }_{ g }\), so the angle  of minimum deviation decreases in water

Increase in focal length of eye lens when eye is fully relaxed.

\(\angle i=\angle r=0^{\circ}\)

(a) Reflection and refraction arise through interaction of incident light with the atomic constituents of matter. Atoms may be viewed as oscillators, which take up the frequency of the external agency (light) causing forced oscillations. The frequency of light emitted by a charged oscillator equals its frequency of oscillation. Thus the frequency of scattered light equals the frequency of incident light.
(b) No, Energy carried by a wave depends on the amplitude of the wave, not on the speed of wave propagation.
(c) For a given frequency, intensity of light in the photon crossing an unit area per uni time.

The resultant intensity is given by
\( I_{R}=I_{1}+I_{2}+2 \sqrt{I_{1} I_{2}} \cos \phi, \)
\(\text { where } I_{1}=I, I_{2}=I+\delta I \)
\(At maxima, \cos \phi=1 \)
\( \therefore I_{\max }=I+I+\delta I+2 \sqrt{I(I+\delta I)} \)
\(\therefore I_{\max }=2 I+2 I=4 I \quad(\because \delta I<
\( At minima, \cos \phi=-1 \)
\(I_{\min }=I+I+\delta I-2 \sqrt{I(I+\delta I)} \)
\( I_{\min } =2 I+\delta I-2\left[I^{2}\left(1+\frac{\delta I}{I}\right)\right]^{1 / 2} \)
\( =2 I+\delta I-2 I\left[1+\frac{1}{2} \frac{\delta I}{I}-\frac{1}{8}\left(\frac{\delta I}{I}\right)^{2}+\ldots\right] \)
Neglecting the higher power, we get
\(I_{\min }=\frac{2 I}{8}\left(\frac{\delta I}{I}\right)^{2}=\frac{1}{4} \frac{(\delta I)^{2}}{I}\)

(i) Reflection and refraction arises through interaction of incident light with the atomic constituents of matter. Atoms may be viewed as oscillators, which take up the frequency of the external agency (light) causing forced oscillations.
The frequency, of light emitted by a charged oscillator equals its frequency of oscillation. Thus, the frequency of scattered light equals the frequency of incident light.
(ii) No, the energy carried by a wave depends on the amplitude of the wave, not on the speed of wave propagation.

Because wavelength of light wave is very small and obstacle of comparable size are not present around us.
The angular width of central maxima is given by, \(\theta={{2\lambda}\over{d}}\)
The width of central maxima is reduced to half when slit width is doubled increases with the increase of wavelength.

(i) According to question

Given magnification ,(m) = -2R = -20cm
f = -10cm
i.e., \(\frac { h_{ 2 } }{ h_{ 1 } } =-2=\frac { -v }{ u } \Rightarrow u=\frac { v }{ 2 } v=2u\)
Now using mirror formula
\(\frac { 1 }{ v } +\frac { 1 }{ u } =\frac { 1 }{ f } \Rightarrow \frac { 1 }{ 2u } +\frac { 1 }{ u } =\frac { 1 }{ -10 } \)
\(\frac { 1+2 }{ 2u } =-\frac { 1 }{ 10 } \left( \therefore f=\frac { R }{ 2 } \right) \)
\(\frac { 3 }{ 2u } =-\frac { 1 }{ 10 } \Rightarrow u=\frac { -10\times 3 }{ 2 } =-15cm\)
v = 2xu = 2x-15 = -30cm
Hence the object distance and image distance are -15cm and -30 cm respectively in front of the mirror
(ii) According to mirror formula i.e. \(\frac { 1 }{ v } +\frac { 1 }{ u } =\frac { 1 }{ f } \)
And we know the value of u and f  for a convex mirror are always negative and positive respectively So, the value of v will always be positive it means convex mirror always forms a virtual image.

(a) Light from the sodium lamp passing through the single Polaroid sheet (P₁) does not show any variation in intensity when this sheet is rotated. However, if the light, transmitted by P₁ is made to pass through another Polaroid sheet (P₂) the light intensity, coming out of P₂ varies from a maximum to zero, and again to maximum, when P₂ is rotated 
These observations are consistent only with the transverse nature of light waves
(b) Intensity of light transmitted through
P₁ = I₀/2
Intensity of light transmitted through
P3 = (I₀/2) x cos² 30°
= 3I₀/8
Intensity of light transmitted through
P₂ = P₃ x cos² (90° - 30°) = 3/8 I₀cos² 60° 
 = 3/32 I₀

\(Here,\ \lambda =555nm=555\times { 10 }^{ -9 }m\)
\( D=25cm=25\times { 10 }^{ -2 }m; \ d\theta =?\)
\(d\theta =\frac { 1.22\lambda }{ D } =\frac { 1.22\times 555\times { 10 }^{ -9 } }{ 25\times { 10 }^{ -2 } }\)
\( =2.7\times { 10 }^{ -6 }rad\)

Let I₀ be the intensity of polarised light after passing through the first polariser P₁. Then the intensity of light after passing through second polariser P₂ will be
I = I₀ cos ² \(\theta\) 
where \(\theta\) is the angle between pass axes of P₁ and P₂. Since P₁ and P₃ are crossed the angle between the pass axes of P₂ and P₃ will be ( \(\pi\) / 2 –\(\theta\) ). Hence the intensity of light emerging from P₃ will be
\(I=I_{0} \cos ^{2} \theta \cos ^{2}\left(\frac{\pi}{2}-\theta\right)\)
\(=I_{0} \cos ^{2} \theta \sin ^{2} \theta=\left(I_{0} / 4\right) \sin ^{2} 2 \theta\)
Therefore, the transmitted intensity will be maximum when \(\theta\) = \(\pi\) / 4.

(a) The shape of the wavefront in case of a light diverging from a point source is spherical. The wavefront emanating from a point source is shown in the given figure.

(b) The shape of the wavefront in case of a light emerging out of a convex lens when a point source is placed at its focus is a parallel grid. This is shown in the given figure.

(c) The portion of the wavefront of light from a distant star intercepted by the Earth is a plane.

\(Here, \ \lambda =589 \ nm,\ c=3\times { 10 }^{ 8 }m/s, \ \mu =1.33\)
(a) For reflected light
\(wavelength,\ \lambda =589\quad nm=589\times { 10 }^{ -9 }m,\quad v=\frac { c }{ \lambda } =\frac { 3\times { 10 }^{ 8 } }{ 589\times { 10 }^{ -9 } } =5.09\times { 10 }^{ 14 }hertz\)
\(speed,\ v=c=3\times { 10 }^{ 8 }m/s\)
(b) For refracted light \( \lambda '=\frac { \lambda }{ \mu } =\frac { 589\times { 10 }^{ -9 } }{ 1.33 } =4.42\times { 10 }^{ -7 }m\)
As frequency remains unaffected on entering another medium,
\(\\ therefore,\quad v'=v=5.09\times { 10 }^{ 14 }hertz\)
\(speed, \ v'=\frac { c }{ \mu } =\frac { 3\times { 10 }^{ 8 } }{ 1.33 } =2.25\times { 10 }^{ 8 }m/s\)

From lens maker's formula,
\(\frac { 1 }{ f } =(\mu -1)\left( \frac { 1 }{ R_{ 1 } } -\frac { 1 }{ R_{ 2 } } \right) \)
where \(\mu =\frac { \mu _{ 2 } }{ \mu _{ 1 } } \)
If \(\mu _{ 2 }=\mu _{ 1 }, \ \frac { 1 }{ f } =0 \ or \ f=\infty \)
\(\therefore \) The lens in the liquid will act like a plane sheet of glass, when refractive index of the lens and the surrounding medium is the same. Therefore, refractive index of surrounding medium,
\(\mu _{ 2 }=\mu _{ 1 }=1.5\)
This liquid medium is not water because refractive index of water=1.33.

(i) Merits of Huygens Wave theory of light:
(a) Wave theory correctly predicted that velocity of light in an optically denser medium is less than that in the rarer medium which is in agreement with the experimental results.
(b) On the basis of wave theory phenomenon of reflection, refraction, interference, diffraction, polarization of light could be explained.
Demiritssof Huygens wave theory of light
(a) Huygens wave theory assumes the existence of luminiferous ether. However,experimentally it couldn't be proved.
(b) This theory couldn't explain rectilinear propagation of light.
(ii) (1) Each pointon a given primary wavefront acts as a source of secondary wavelets, sending out disturbances (waves) in all directions in a similar manner as the original source of light does.
(2) The new position of the wavefront at any instant (secondary wavefront) is given by the forward envelope to the secondary wavelets at that instant.
Huygens' construction

Using this principle the laws of reflection and refraction can be verified.
(iii) Primary source of light: It is a real source of light. It generates light itself and sends primary wavefronts in all directions.
Secondary source of light: It is a fictitious source of light presents on the wavefront and sends out secondary waves only in forward direction.