As shown in the figure, in Young's double slit experiment, a thin plate of thickness t=10 \mu m and refractive index \mu=1.2 is inserted infront of slit S_1 . The experiment is conducted in air (\mu=1) and uses a monochromatic light of wavelength \lambda=500 ~nm . Due to the insertion of the plate, central maxima is shifted by a distance of x \beta_0 . \beta_0 is the fringe-width befor the insertion of the plate. The value of the x is \underline\hspace2cm .

Wave Optics PYQ — JEE Main 2023 · 1 Feb · Shift 2 · Q67 with Solution

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Correct answer: 4

When a thin plate of thickness $t$ and refractive index $\mu$ is placed in front of one slit in YDSE, the additional optical path introduced is

\Delta = (\mu-1)t

Given:

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

\mu = 1.2

So,

\Delta = (1.2-1)(10\,\mu\text{m}) = 0.2\times 10\,\mu\text{m} = 2\,\mu\text{m}

In YDSE, if an extra path difference $\Delta$ is introduced in one path, the fringe pattern shifts by

\text{shift} = \frac{\Delta}{\lambda}\,\beta_0

Hence if the shift is $x\beta_0$ , then

x = \frac{\Delta}{\lambda}

Given:

\lambda = 500\text{ nm} = 0.5\,\mu\text{m}

Therefore,

x = \frac{2\,\mu\text{m}}{0.5\,\mu\text{m}} = 4

\boxed{x=4}

The derived answer matches the stored correct answer.