Optics Question 1140
A double slit experiment is performed with light of wavelength 500 nm. A thin film of thickness 2 micrometers and refractive index 1.5 is introduced in the path of the upper beam. The location of the central maximum will
[AIIMS 2003]
Options:
A) Remain unshifted
B) Shift downward by nearly two fringes
C) Shift upward by nearly two fringes
D) Shift downward by 10 fringes
Show Answer
Answer:
Correct Answer: C
Solution:
If shift is equal to n fringes width, then $ n=\frac{(\mu -1)t}{\lambda }=\frac{(1.5-1)\times 2\times {{10}^{-6}}}{500\times {{10}^{-9}}}=\frac{1}{500}\times 10^{3}=2 $ Since a thin film is introduced in upper beam, the shift will be upward.
BETA
