SATHEE: Optics Question 142

Optics Question 142

Question: In an interference arrangement similar to Young’s double slit experiment, the slits S1 and S2 are illuminated with coherent microwave sources each of frequency 106 Hz. The sources are synchronized to have zero phase difference. The slits are separated by distance d = 150 m. The intensity I $(θ)$ is measured as a function ofq, where q is defined as shown. If I0 is maximum intensity, then $I (θ)$ for $0 \leq θ \leq 90^{o}$ is given by

[IIT 1995]

Options:

A) $I (θ) = I_{0}$ for $θ = 0^{o}$

B) $I (θ) = I_{0} / 2$ for $θ = 30^{o}$

C) $I (θ) = I_{0} / 4$ for $θ = 90^{o}$

D) $I (θ)$ is constant for all values of q

Show Answer

Answer:

Correct Answer: B

Solution:

For microwave $λ = \frac{c}{f} = \frac{3 \times 10^{8}}{10^{6}} = 300 m$

As $Δ x = d \sin θ$

Phase difference $φ = \frac{2 π}{λ}$

(Path difference) $= \frac{2 π}{λ} (d \sin θ) = \frac{2 π}{300} (150 \sin θ) = π \sin θ$

$I_{R} = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2}} \cos φ$

Here $I_{1} = I_{2}$ and $φ = π \sin θ$

$∴ I_{R} = 2 I_{1} [1 + \cos (π \sin θ)] = 4 I_{1} \cos^{2} (\frac{π \sin θ}{2})$

IR will be maximum when $\cos^{2} (\frac{π \sin θ}{2}) = 1$

$∴ {(I_{R})}_{max} = 4 I_{1} = I_{o}$

Hence $I = I_{o} \cos^{2} (\frac{π \sin θ}{2})$

If $θ = 0,$ then $I = I_{o} \cos θ = I_{o}$

If $θ = 30^{\circ},$

then $I = I_{o} \cos^{2} (π / 4) = I_{o} / 2$

If $θ = 90^{o},$ then $I = I_{o} \cos^{2} (π / 2) = 0$

Optics Question 142

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Optics Question 142

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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