Resolving power of a microscope is the reciprocal of the minimum separation of two points seen as distinct
Resolving power \(=\frac{1}{d_{\min }}=\frac{2 \mu \sin \beta}{1.22 \lambda}\)
(i) As resolving power ∝ μ, on increasing the refractive index of the medium between the object and objective lens, resolving power increases.
(ii) As resolving power \(\propto \frac{1}{\lambda},\) on increasing the wavelength of the radiation resolving power decreases.

As, fringe width,
\(\beta =\frac { 2\lambda }{ 2d } \)
\(\therefore \ 7.2mm=\frac { D\times 660nm }{ 2d } \)
\(and \ 8.1mm=\frac { D\times \lambda }{ 2d } \)
Dividing (i) by (ii)
\(\therefore \)  \(\frac { 7.2mm }{ 8.1mm } =\frac { 640nm }{ { \lambda }^{ ' } } \) 
or \({ \lambda }^{ ' }\)= 640 x \(2.1\over7.2\)nm = 720 nm
Also \({ y }_{ n }={ n }_{ 1 }\frac { D\lambda }{ 2d } ={ n }_{ 2 }\frac { D{ \lambda }^{ ' } }{ 2d } \)
\(or \ { n }_{ 1 }\lambda = \ { n }_{ 2 }{ \lambda }^{ ' }\)
\(\Rightarrow { n }_{ 1 }\times 640={ n }_{ 2 }\times 720\)
\(\frac { { n }_{ 1 } }{ { n }_{ 2 } } =\frac { 720 }{ 640 } =\frac { 9 }{ 8 } \)

(a) (i) Ray of light should travel from denser to rarer medium.
(ii) Angle of incidence should be more than the critcical angle 
(b) \(\mu =\frac { 1 }{ { sini }_{ c }\quad } \) where i_c is the critical angle

Depth of focus indicates the range of the object distances over which the focussing is reasonably good.

(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.

(a) The required graph would have the form shown as:

Using I₂ = I₁cos²\(\theta\)
(b) I₁ = Light transmitted by P₁ 
I₃ = Light transmitted by P₃ = I₁cos² \(\beta\)
I₂ = Light transmitted by P₂ = 1₃cos² (\(\theta\)-\(\beta\))
Alternatively, (Award mark to student who indicates current value of  I₁, I₂ and I₃ by making a diagram)   
I₂ = I₃
I₁ COS² \(\beta\).cos² (\(\theta\)-\(\beta\) ) = I₁ COS² \(\beta\)    
\(\theta\)= \(\beta\) 
Also I₁ = I₂
I₁ = I₁ cos² \(\beta\)         
or cos² \(\theta\) = 1
\(\therefore\)  \(\theta\) = 0⁰   
∴ \(\beta\) = 0⁰  

Here, \(a=4 \ cm=4\times { 10 }^{ -2 }m\) 
\(\\ \lambda =2\quad cm=2\times { 10 }^{ -2 }m\)
Angular spread of central maximum \((2\theta )\) is
\(2\theta =\cfrac { 2\lambda }{ a } =\cfrac { 2\times 2\times { 10 }^{ -2 } }{ 4\times { 10 }^{ -2 } } =1 \ rad\)

Here, f₀ = 0.3m and f_e = 3cm, \(\alpha\)= 0.5⁰, \(\beta\)=?
Use \({\beta\over\alpha}={f_0\over f_e}\)

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, slit width,d = 0.28 mm = \(0.28\times { 10 }^{ -3 }m\)
Distance between slit and screen, D = 1.4m
 \(y=1.2cm=1.2\times { 10 }^{ -2 }m,n=4,\lambda =?\)
For constructive interference,
 \(y=\eta \lambda \frac { D }{ d } or\lambda =\frac { yd }{ nD }\)
\( \\ =\frac { 1.2\times { 10 }^{ -2 }\times 0.28\times { 10 }^{ -3 } }{ 4\times 1.4 } =6\times { 10 }^{ -7 }m\)
= 600 nm
Hence, the wavelength of the light is 600 nm.

\(Here, \ \lambda =589 \ nm,\ c=3\times { 10 }^{ 8 }m/s, \ \mu =1.33\)
(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,
\(\\ therefore,\quad v'=v=5.09\times { 10 }^{ 14 }hertz\)
\(speed, \ v'=\frac { c }{ \mu } =\frac { 3\times { 10 }^{ 8 } }{ 1.33 } =2.25\times { 10 }^{ 8 }m/s\)

Here m =-50,  L = 102cm
As \(m=-{f_0\over f_e}=-50, f_0=50f_e\)
Also, L = f₀ + f_e= 102;
50f₀ + f_e= 102, f_e = 2 cm
f₀ = 50 f_e= 50 x 2= 100 cm
\(P_o={100\over f_0}={100\over100}\)= 1D,
\(P_e={100\over f_e}={100\over2}\)= 50D

(c): The waves diffracted from the edges of circular obstacle, placed in the path of light, interfere constructively at the centre of the shadow resulting in the formation of a bright spot.

(a): For reflected system of the film, the maxima or constructive interference is\(2 \mu t \cos r\) = while the maxima for transmitted system of film is given by equation
\(2 \mu t \cos r=n \lambda\)
where t is thickness of the film and r is angle of refraction.
From these two equations we can see that condition for maxima in reflected system and transmitted system are just opposite.

(c): The reflection and refraction of light occurs on account of interaction of light with the atoms of the surface of separation. These atoms can be regarded as oscillators. Light incident on the interface forces the atomic oscillators to oscillate with frequency of incident light. As frequency oflight emitted by these (charged). oscillators is equal to their own frequency of oscillation, therefore, reflected and refracted light have the same frequency as that of incident light.

(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.