NEET Solved Paper 2016 Question 18
Question: The intensity at the maximum in a Young’s double slit experiment is $ {I_0}\text{.} $ Distance between two slits is $ d=5\lambda , $ where $ \lambda $ the wavelength of light is used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance D = 10 d?
Options:
A) $ I _0 $
B) $ \frac{I _0}{4} $
C) $ \frac{3}{4}I _0 $
D) v
Show Answer
Answer:
Correct Answer: D
Solution:
Path difference $ =S _2P-S _1P $ $ =\sqrt{D^{2}+d^{2}}-D $ $ =D( 1+\frac{1}{2}\frac{d^{2}}{D^{2}} )-D $ $ =D[ 1+\frac{d^{2}}{2D^{2}}-1 ]=\frac{d^{2}}{2D} $ $ \Delta x=\frac{d^{2}}{2\times 10d}=\frac{d}{20}=\frac{5\lambda }{20}=\frac{\lambda }{4} $ $ \Delta ,o|,=\frac{2\pi }{\lambda }.\frac{\lambda }{4}=\frac{\lambda }{2} $
So, intensity at the desired point is $ I=I _0{{\cos }^{2}}\frac{\text{o }|}{2}=I _0{{\cos }^{2}}\frac{\pi }{4}=\frac{I _0}{2} $
BETA
