Wave Optics

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.

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

Explanation- when polariser is not used

A=A_perp+A

letA₁= asinwt and A₂=asin(wt+ $\varnothing$ )

now superposition principle for perpendicular polariser

A_R= asinwt+ asin(wt+ $\varnothing$ )

A_R=a(2cos $\varnothing /2$ sin(wt+ $\varnothing$ ))

A_R=2acos $\varnothing /2$  sin(wt+ $\varnothing$ )

This eqn is also same for parallel polariser

A_R=2acos $\varnothing /2$  sin(wt+ $\varnothing$ )

And we know that intensity is directly proportional to square of amplitude

(A_R)²= (A_perp)²+(A)²

So resultant intensity is

I=4(a)²cos^{2 $\varnothing /2\frac{1}{T}{\int }_0^T{sin}^2(wt+\varnothing )$} dt + 4(a)²cos^{2 $\varnothing /2\frac{1}{T}{\int }_0^T{sin}^2(wt+\varnothing )$} dt

I= 8(a)²cos^{2 $\varnothing /2$} (1/2)                                     (after integrating sin value we got 1/2)

I= 4(a)²cos^{2 $\varnothing /2$}

So I= 4(a)²

But when there is a polariser

Perpendicular polariser is blocked

I=2(a)²cos^{2 $\varnothing /2\frac{1}{T}{\int }_0^T{sin}^2(wt+\varnothing )$} dt + (a)^{2 $\frac{1}{T}{\int }_0^T{sin}^2\left(wt\right)$}

I=2(a)²cos^{2 $\varnothing /2+$} 1/2(a)²

When $\varnothing =0$

I_max=2(a)²+1/2(a)²

I_max=5/2(a)²

I_max=5/2(I/4)=5I/8

At first minima it becomes I/8

10.17 If the width of the slit is made double the original width, then the size of the central diffraction band reduces to half and the intensity increases up to four times.

The interference pattern in a double-slit experiment is modulated by diffraction from each slit. The pattern is the result of the interference of the diffracted wave from each slit.

This is because light waves are diffracted from the edge of the circular obstacle, which interferes constructively at the centre of the shadow. This constructive interference produces a bright spot.

Bending of waves by obstacles by a large angle is possible when the size of the obstacle is comparable to the wavelength of the waves.

Wavelength of light waves is too small in comparison to the size of the obstacle. Thus the diffraction angle will be small, resulting in students cannot see each other. On the other hand, the size of the wall is comparable to the wavelength of the sound waves, bending of waves takes place at large angle – Resulting in students can hear each other.

The justification is that in ordinary optical instruments, the size of the aperture involved is much larger than the wavelength of the light used.