Optics Question 466
Question: In the ideal double-slit experiment, when a glass plate (refractive index 1.5) of thickness $ t $ is introduced in the path of one of the interfering beams (wavelength $ \lambda $ ), the intensity at the position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass plate is
Options:
A) $ 2\lambda $
B) $ 2\lambda /3 $
C) $ \lambda /3 $
D) $ \lambda $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Path difference due to slab should be integral multiple of $ \lambda $ . Hence, $ \Delta x=n\lambda $ Or $ (\mu -1)t=n\lambda , $
$ n=1,2,… $ Or $ t=\frac{n\lambda }{\mu -1} $ For minimum value of t, n=1.
$ \therefore t=\frac{n\lambda }{\mu -1}=\frac{\lambda }{1.5-1}=2\lambda $
BETA
