SATHEE: Optics Question 144

Optics Question 144

Question: In Young’s double slit experiment, the intensity on the screen at a point where path difference is l is K. What will be the intensity at the point where path difference is $λ / 4$

[RPET 1996]

Options:

A) $\frac{K}{4}$

B) $\frac{K}{2}$

C) K

D) Zero

Show Answer

Answer:

Correct Answer: B

Solution:

By using phase difference $φ = \frac{2 π}{λ} (Δ)$

For path differencel, phase difference $φ_{1} = 2 π$ and for path difference l/4, phase difference f2 = p/2. Also by using $I = 4 I_{0} \cos^{2} \frac{φ}{2}$

Therefore $\frac{I_{1}}{I_{2}} = \frac{\cos^{2} (φ_{1} / 2)}{\cos^{2} (φ_{2} / 2)}$

Therefore $\frac{K}{I_{2}} = \frac{\cos^{2} (2 π / 2)}{\cos^{2} (\frac{π / 2}{2})} = \frac{1}{1 / 2}$

Therefore $I_{2} = \frac{K}{2} .$

Optics Question 144

Question: In Young’s double slit experiment, the intensity on the screen at a point where path difference is l is K. What will be the intensity at the point where path difference is $λ / 4$

Options:

Answer:

Solution:

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Optics Question 144

Question: In Young’s double slit experiment, the intensity on the screen at a point where path difference is l is K. What will be the intensity at the point where path difference is λ/4

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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Question: In Young’s double slit experiment, the intensity on the screen at a point where path difference is l is K. What will be the intensity at the point where path difference is $λ / 4$