Question

A double-slit interference pattern is created by two narrow slits spaced 0.21 mm apart. The distance between the first and the fifth minimum on a screen 61 cm behind the slits is 6.2 mm What is the wavelength of the light used in this experiment?

Answer 1

Answer:

$\lambda =533.6 nm$

Explanation:

the slits spacing, d = 0.21 mm

distance of screen, D = 61 cm

The condition for minima is given as

$dsin(\theta) = \left ( n+\frac{1}{2} \right )\lambda$

So, first minima, n = 0

$dsin(\theta_1) = \frac{1}{2}\lambda$

fifth minima, n = 4

$dsin(\theta_5) = \frac{9}{2}\lambda$

$d(sin(\theta_5) -sin(\theta_1))= 4\lambda$

For small angle

$d(tan(\theta_5) -tan(\theta_1))= 4\lambda$

From the figure:

$d(\frac{y_5}{D}-\frac{y_1}{D})= 4\lambda$

$\frac{d}{D} (y_5-y_1) = 4\lambda$

$\lambda = \frac{d}{4D} (y_5-y_1)$

$\lambda = \frac{0.021}{4(61)} (6.2 \times 10^{-3})$

$\lambda =533.6 nm$