SATHEE: Optics Question 853

Optics Question 853

Question: Interference fringes were produced using white light in a double slit arrangement. When a mica sheet of uniform thickness of refractive index 1.6 (relative to air) is placed in the path of light from one of the slits, the central fringe moves through some a distance. This distance is equal to the width of 30 interference bands if light of wavelength is used. The thickness $(i n μ m)$ of mica is

Options:

A) 90

B) 12

C) 14

D) 24

Show Answer

Answer:

Correct Answer: D

Solution:

[d] Shift of fringe pattern $= (μ - 1) \frac{t D}{d}$

$∴ \frac{30 D (4800 \times 10^{- 10})}{d} = (0.6) t \frac{D}{d}$

$30 \times 4800 \times 10^{- 10} = 0.6 t$

$t = \frac{30 \times 4800 \times 10^{- 10}}{0.6} = \frac{1.44 \times 10^{- 5}}{0.6} = 24 \times 10^{- 6}$

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Optics Question 853

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