How does the fringe width of interference fringes change, when the whole apparatus of Young's experiment is kept in a liquid of refractive index 1.13?

Solution

Fringe width is directly proportional to wavelength and inversely proportional to refractive index.

The formula is given by,

In air, Fringe width, x = $\frac{D λ}{d}$

In medium, $Fringe width, x = \frac{Dλ}{μd}$
where,
d is the distance between the slits and,
D is the distance between the screen and the slit.

Therefore, fringe width decreases by a factor of $μ (1.33)$ when placed in the liquid.

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Wave Optics

How does the fringe width of interference fringes change, when the whole apparatus of Young's experiment is kept in a liquid of refractive index 1.13?

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