In a Young's double-slit experiment…

In a Young's double-slit experiment the slit separation is 0.50 mm, the screen is 1.5 m away and monochromatic light of wavelength 600 nm is used. A…

One or more options correct

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

0.50\ \mathrm{mm}

, the screen is

1.5\ \mathrm{m}

away and monochromatic light of wavelength

600\ \mathrm{nm}

is used. A thin transparent sheet of refractive index

1.5

and thickness

12\ \mu\mathrm{m}

is placed in front of one slit. Which of the following statements is/are correct?

A. The central bright fringe shifts by $18\ \mathrm{mm}$ towards the side of the sheet.
B. The fringe width increases because the sheet makes the optical path longer.
C. The point on the screen that was originally the centre of the pattern now becomes a dark fringe.
D. If the whole apparatus (with the sheet removed) is immersed in water of refractive index $4/3$ , the fringe width becomes $2.4\ \mathrm{mm}$ .

Show answer and solution

Answer: A

The sheet adds an extra optical path

(\mu-1)t=0.5\times12\ \mu\mathrm{m}=6\ \mu\mathrm{m}

in one beam, so the zero-order fringe moves by

\Delta y=\dfrac{(\mu-1)tD}{d}=\dfrac{6\times10^{-6}\times1.5}{0.5\times10^{-3}}=18\ \mathrm{mm}

towards the sheet (A true). The fringe width

\beta=\dfrac{\lambda D}{d}=1.8\ \mathrm{mm}

is determined only by

\lambda

D

and

d

, and is unaffected by the sheet (B false). The extra path equals

\dfrac{6\ \mu\mathrm{m}}{0.6\ \mu\mathrm{m}}=10

wavelengths, i.e. an integer number, so the original central position still has a path difference of

10\lambda

and remains bright (C false). Immersion in water reduces the wavelength to

\lambda/\mu_w

, so

\beta'=1.8\times\tfrac34=1.35\ \mathrm{mm}

, not

2.4\ \mathrm{mm}

(D false).

Original question written in the JEE Advanced 2016 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake