Monochromatic light of wavelength 500 ~nm is used in Young's double slit experiment. An interference pattern is obtained on a screen. When one of the slits is covered with a very thin glass plate (refractive index =1.5 ), the central maximum is shifted to a position previously occupied by the 4^ th bright fringe. The thickness of the glass-plate is \underline\hspace2cm \mu m .

Wave Optics PYQ — JEE Main 2024 · 9 Apr · Shift 2 · Q89 with Solution

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Correct answer: 4

When a thin glass plate of thickness $t$ and refractive index $\mu$ is inserted in front of one slit, the optical path in that arm increases by

\Delta = (\mu-1)t

This causes the entire interference pattern to shift.

The new central maximum shifts to the position previously occupied by the $4^{\text{th}}$ bright fringe.

That means the path difference introduced by the plate is equal to $4\lambda$ :

(\mu-1)t = 4\lambda

Given:

\lambda = 500\,\text{nm} = 500 \times 10^{-9}\,\text{m}

\mu = 1.5

So,

(1.5-1)t = 4(500 \times 10^{-9})

0.5t = 2000 \times 10^{-9}

t = \frac{2000 \times 10^{-9}}{0.5}

t = 4000 \times 10^{-9}\,\text{m}

t = 4 \times 10^{-6}\,\text{m}

t = 4\,\mu\text{m}

The thickness of the glass plate is

\boxed{4}

in micrometres.