Optics Question 1140

Question: A double slit experiment is performed with light of wavelength 500 nm. A thin film of thickness 2 mm 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. So shift will be upward.



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