In a double slit experiment the distance between the slits is 0.1 cm and the screen is placed at 50 cm from the slits plane. [2026]

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Solution

Q.
In a double slit experiment the distance between the slits is 0.1 cm and the screen is placed at 50 cm from the slits plane. When one slit is covered with a transparent sheet having thickness t and refractive index n (= 1.5), the central fringe shifts by 0.2 cm. The value of t is __________ cm. [2026]

1 $6.0 \times 10^{- 3}$

2 $5.6 \times 10^{- 4}$

3 $8 \times 10^{- 4}$

4 $8 \times 10^{- 4}$

Ans.
(3)

$d \sin θ = (μ - 1) t$

$d [\frac{x}{D}] = (μ - 1) t$

$t = \frac{x d}{D (μ - 1)}$

$= \frac{(0.2) (0.1)}{50 (1.5 - 1)}$

$t = 8 \times 10^{- 4} cm$

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