SATHEE: Optics 6 Question 37

Optics 6 Question 37

39. In an interference arrangement similar to Young’s double-slit experiment, the slits $S_{1}$ and $S_{2}$ are illuminated with coherent microwave sources, each of frequency $10^{6} H z$ . The sources are synchronized to have zero phase difference. The slits are separated by a distance $d = 150.0 m$ . The intensity $I (θ)$ is measured as a function of $θ$ , where $θ$ is defined as shown. If $I_{0}$ is the maximum intensity, then $I (θ)$ for $0 \leq θ \leq 90^{\circ}$ is given by

$(1995, 2 M)$

(a) $I (θ) = I_{0} / 2$ for $θ = 30^{\circ}$

(b) $I (θ) = I_{0} / 4$ for $θ = 90^{\circ}$

(d) $I (θ)$ is constant for all values of $θ$

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Solution:

The intensity of light is $I (θ) = I_{0} \cos^{2} \frac{δ}{2}$

where, $δ = \frac{2 π}{λ} (Δ x) = \frac{2 π}{λ} (d \sin θ)$

(a) $For θ = 30^{\circ}$

$\begin{aligned} λ = \frac{c}{ν} = \frac{3 \times 10^{8}}{10^{6}} = 300 m and d = 150 m \\ ∴ & \frac{δ}{2} = \frac{2 π}{300} (150) \frac{π}{2} = \frac{π}{2} \\ ∴ & I (θ) = I_{0} \cos^{2} \frac{π}{4} = \frac{I_{0}}{2} \end{aligned}$

(b) For $θ = 90^{\circ}$

$δ = \frac{2 π}{300} (150) (1) = π$

$\frac{δ}{2} = \frac{π}{2} and I (θ) = 0$

$∴ I (θ) = I_{0}$

[option (c)]

Optics 6 Question 37

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Optics 6 Question 37

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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