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MHT CET 2026 (13 May, Shift 2) · YDSE: Film, Medium & Two Wavelengths

YDSE: Film, Medium & Two Wavelengths question from MHT CET 2026 (13 May, Shift 2)

Q9MHT CET 2026 · 13 MAY · SHIFT 2Medium

YDSE: Film, Medium & Two Wavelengths

A double slit experiment is performed with light of wavelength $500$ nm. A thin film of thickness $2\ \mu$ m and refractive index $1.5$ is introduced in the path of the upper beam. The location of the central maximum

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Answer: (c)

Number of fringes shifted $n = \frac{(\mu-1)t}{\lambda} = \frac{0.5\times2\times10^{-6}}{500\times10^{-9}} = 2$ .
The central maximum shifts towards the side of the film, i.e. upward.
So the answer is (c) shift upward by nearly two fringes.

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