NEET Solved Paper 2014 Question 36
Question: In the Young’s double-slit experiment, the intensity of light at a point on the screen (where the path difference is $ \lambda $ ) is K, ( $ \lambda $ being the wavelength of light used). The intensity at a point where the path difference is $ \lambda $ , / 4, will be [AIPMT 2014]
Options:
A) K
B) K/4
C) K/2
D) zero
Show Answer
Answer:
Correct Answer: C
Solution:
- For net intensity $ l’=4l _0{{\cos }^{2}}\frac{\phi }{2}( \phi =\frac{2\pi }{\lambda }\times \lambda ) $ For the first case, $ K=4l _0{{\cos }^{2}}[\pi ] $ $ K=4l _0…..(i) $ For the second case $ K’=4l _0{{\cos }^{2}}( \frac{\pi /2}{2} )( \phi =\frac{2\pi }{\lambda }\times \frac{\lambda }{4} ) $ $ 4l _0{{\cos }^{2}}(\pi /2) $ $ K’=2l _0…..(ii) $ Comparing Eqs. (i) and (ii) $ K’=K/2 $
