Question

g When a laser light of unknown wavelength is passed through two narrow slits separated by 150 μm, an interference pattern is observed on a screen 1.5 m away. The distance between the central spot and first order constructive interference is found to be 5 mm. a) Find the wavelength of the laser light, in nanometers. (b) At what angle, in degrees, from the central spot is the second-order constructive interference formed?

Answer 1

Answer:

(a) 500 nm

(b) $\theta = 0.382^{\circ}$

Solution:

Slit separation, d = $150\mu m = 150\times 10^{- 6}\ m$

Distance from the screen, x = 1.5 m

Distance between central spot and 1st order constructive interference, $y_{1} = 5 mm = 5\times 10^{- 3}\ m$

Now,

(a) To calculate the wavelength of laser light:

From Double-slit experiment, we know that the distance between the central fringe and the nth bright fringe is given by:

$y_{n} = \frac{n\lambda x}{d}$

where

n = 1

$\lambda$ = wavelength of the light used

Thus

$y_{1} = \frac{1.\lambda x}{d}$

$5\times 10^{- 3} = \frac{\lambda \times 1.5}{150\times 10^{- 6}}$

$\lambda = 5\times 10^{- 7}\ m = 500\ nm$

(b) To calculate the angle for the formation of the constructive interference of second order:

Here

n = 2

Thus

$y_{2} = \frac{2\lambda x}{d}$

$tan\theta = \frac{\frac{2\lambda x}{d}}{x} = \frac{2\lambda }{d}$

$tan\theta = \frac{2\times 500\times 10^{- 9}}{150\times 10^{- 6}}\ m$

$\theta = tan^{- 1} (6.67\times 10^{- 3})$

$\theta = 0.382^{\circ}$