Optics Question 144
In Young’s double slit experiment, the intensity on the screen at a point where path difference is λ 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} $
K
D) Zero
Show Answer
Answer:
Correct Answer: B
Solution:
By using phase difference $ \varphi =\frac{2\pi }{\lambda }(\Delta x) $
For path difference l, phase difference $ {\varphi _{1}}=2\pi $ and for path difference l/4, phase difference $ {\varphi _{2}}=\pi/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} $.
BETA
