Wave Optics — Practice

Question 1

In YDSE, lambda = 600 nm, d = 0.2 mm, D = 1.5 m. Find the fringe width.

Accepted Answer

beta = lambda D/d = (600e-9 x 1.5)/(0.2e-3) = 4.5e-3 m = 4.5 mm.

Question 2

A single slit of width 0.1 mm with lambda = 500 nm and screen at 1 m. Find the central maximum width.

Accepted Answer

w = 2 lambda D / a = 2 x 500e-9 x 1 / 0.1e-3 = 1e-2 m = 1 cm.

Question 3

Find Brewster's angle for glass (mu = 1.5).

Accepted Answer

tan(theta_B) = 1.5, so theta_B = arctan(1.5) = 56.3 degrees.

Question 4

Two polarisers are at 60 degrees. If I0 falls on the first, find the transmitted intensity.

Accepted Answer

After the first polariser I1 = I0/2. After the second I2 = I1 cos^2(60) = (I0/2)(1/4) = I0/8.

Question 5

What happens to YDSE fringe width if the apparatus is immersed in water (mu = 4/3)?

Accepted Answer

Wavelength becomes lambda/mu = 3 lambda/4, so the fringe width decreases to 3/4 of its original value.

Practice — Wave Optics