JEE Main 2020 · Physics · Waves and Sound · hard

In a Young's double slit experiment, the slit separation is 0.5 mm, the screen is 1 m away and λ = 500 nm. The maximum intensity is Imax. A point P…

Single correct option

In a Young's double slit experiment, the slit separation is

0.5\ \mathrm{mm}

, the screen is

1\ \mathrm{m}

away and

\lambda = 500\ \mathrm{nm}

. The maximum intensity is

I_{max}

. A point

P

on the screen is the position of the third bright fringe from the centre. A thin transparent sheet of refractive index

1.5

and thickness

2.25\ \mu\mathrm{m}

is now placed in front of one slit. The intensity at

P

becomes

Show answer and solution

Answer: (A) $\dfrac{I_{max}}{2}$

Originally the path difference at

P

3\lambda

(check:

y = 3\beta = 3\ \mathrm{mm}

and

dy/D = 1.5\ \mu\mathrm{m}

). The sheet adds an optical path

(\mu - 1)t = 0.5\times2.25 = 1.125\ \mu\mathrm{m} = 2.25\lambda

to one beam, so the net path difference becomes

3\lambda \pm 2.25\lambda = 0.75\lambda

5.25\lambda

(depending on which slit is covered). Either way the phase difference is

\phi = 2\pi\Delta/\lambda = 1.5\pi

10.5\pi

, and for equal amplitudes

I = I_{max}\cos^2(\phi/2) = I_{max}\cos^2(3\pi/4) = I_{max}/2

Original question written in the JEE Main 2020 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake