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 $ \lambda /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 $ \varphi =\frac{2\pi }{\lambda }(\Delta ) $

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

Therefore $ \frac{I _{1}}{I _{2}}=\frac{{{\cos }^{2}}({\varphi _{1}}/2)}{{{\cos }^{2}}({\varphi _{2}}/2)} $

Therefore $ \frac{K}{I _{2}}=\frac{{{\cos }^{2}}(2\pi /2)}{{{\cos }^{2}}( \frac{\pi /2}{2} )}=\frac{1}{1/2} $

Therefore $ I _{2}=\frac{K}{2}. $



जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक