In a Youngs double slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. When the entire apparatus is immersed in a liquid, the fourth bright fringe shifts to a position 1.0 cm from the central fringe. Calculate the refractive index of the liquid.

A. 1.2 — The fringe width is inversely proportional to the refractive index of the medium. Since the position of the fourth fringe is proportional to the fringe width, the ratio of the positions equals the inverse ratio of the refractive indices, leading to mu = 1.2/1.0 = 1.2.

JEE Advanced · PhysicsHard

$In a Youngs double slit experiment, the slits are separated by and the screen is placed away. The distance between the central bright fringe and the fourth bright fringe is . When the apparatus is immersed in a liquid, the fourth bright fringe shifts to . Calculate the refractive index of the liquid.$

A. $1.2$
B. $1.44$
C. $1.25$
D. $1.33$

Show correct answer & step-by-step solution

Correct answer: A — $1.2$

Solution

The fringe width β = \frac{λ D}{d} becomes β^{'} = \frac{λ D}{μ d} in a medium. Thus, y_{4} / y_{4}^{'} = μ, which gives μ = 1.2/1.0 = 1.2 .

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