Wave Optics

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- As per the given question, monochromatic light emerging from polaroid (I) is plane polarised. When polaroid (II) is placed infront of this polaroid (I), and rotated till no light passes through polaroid (II), then (I) and (II) are set in crossed positions, i.e., pass axes of I and II are at 90°.

Consider the above diagram where a third polaroid (III) is placed between polaroid (I) and polaroid II.

When a third polaroid (III) is placed in between (I) and (II), no light will emerge from (II), if pass axis of (III) is parallel to pass axis of (I) or (II). In all other cases, light will emerge from (II), as pass axis of (II)will no longer be at 90° to the pass axis of (III).

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- refractive index = 1.38 refractive index = 1.5

^{$\lambda =5500A$ 0}

Consider a ray incident at an angle i. A part of this ray is reflected from the air-film interface And apart refracted inside.

This is partly reflected at the film-glass interface and a part transmitted. A part of the

reflected ray is reflected at the film-air interface and a part transmitted as r2 parallel to r 1. Of course successive reflections and transmissions will keep on decreasing the amplitude of the wave. Hence, rays r 1 and r2 shall dominate the behaviour. If incident light is to be transmitted through the lens, r 1 and r2 should interfere destructively. Both the reflections at A and D are from lower to higher refractive index and hence, there is no phase change on reflection. The optical path difference between r₂ and r₁ is

n(AD+CD)-AB

if d is the thickness of the film , then AD=CD=d/cosr

AB=AC sini

AC/2= d tanr

AC= 2dtanr

Hence AB= 2d tanr sini

Thus the optical path difference = $\frac{2nd}{cosr}$ -2dtanrsini

= 2. $\frac{sinid}{sinrcosr}$ - 2d $\frac{sinr}{cosr}sini$

= 2dsin{ $\frac{1-{sin}^{2}r}{sinrcosr}$ }

= 2ndcosr

For path difference $\frac{\lambda }{2}$

2ndcosr= $\frac{\lambda }{2}$

ndcosr= $\frac{\lambda }{4}$

for photographic lenses, the sources are normally in vertical plane

i=r=0⁰

ndcos0= $\frac{\lambda }{4}$

d= $\frac{\lambda }{4n}$ =5500/4 $*1.38$ = 1000A⁰

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation-consider the disturbance at the receiver R₁ which is at a distance d from B

Y_A= acos(wt)  and path difference is $\frac{\lambda }{2}$  hence phase difference is $\pi$ .

Thus the wave R₁ because of B

Y_B= acos(wt- $\pi$ )= - acoswt here path difference is $\lambda$  and hence phase difference is $2\pi$

Thus R₁ because of C

Y_c= acos(wt-2 $\pi$ )= acoswt

(i)let the signal picked up at R₂ from B be Y_B= a₁cos(wt)

The path difference between signal at D and that B is $\frac{\lambda }{2}$

Y_D= -a₁cos(wt)

The path difference between signal at A and that atB is

$\sqrt[]{d^2+{\left(\frac{\lambda }{2}\right)}^2}$ -d = d( ${1+\frac{{\lambda }^2}{4D^2})}^{1/2}$ -d = $\frac{{\lambda }^2}{8d^2}$

$asd?\lambda$  therefore path difference os 0

$phasedifference=\frac{2\pi }{\gamma }\frac{{\lambda }^2}{8d^2}$

_{$Y$ A}=a₁cos(wt- $\varnothing$ )

_{$Y$ C}=a₁cos(wt- $\varnothing$ )

Therefore signal picked up by R₂ is

Y_A+Y_b+Y_c+Y_d= y = 2a₁cos(wt- $\varnothing$ )

|y|² = 4a₁²cos²(wt- $\varnothing$ )

=2a₁². Thus R₁ picks up the larger signal.

(ii) if B is swiched off

R₁ picks up y = acoswt

^{IR1= $\frac{1}{2}$ a2}

R₂ picks up  y = a coswt

I_R2= $\frac{1}{2}$ a²

(iii) thus R₁ and R₂ pick up the same signal

If D is switched off

R₁ picks up by y = acoswt

I_R1= $\frac{1}{2}$ a²

Y= 3acoswt

I_R2=9a²/2

R₂ picks up larger signal compared to R₁

(iv) thus a signal at R₁ indicates B has been switched off and an enhanced signal at R₂ indicates D has been switched off.