Question

Two slits are spaced 0.33mm apart. If the angle between the first dark fringe and the central maximum is 0.055º, what is the wavelength of a light falling on these two slits?

Answer 1

Answer:

$6.33\cdot 10^{-7} m$

Explanation:

The condition for the first minimum (= first dark fringe) in the interference pattern produced by diffraction from a double slit is

$d sin \theta = \frac{\lambda}{2}$

where

$d=0.33 mm = 0.33\cdot 10^{-3}m$ is the separation between the slits

$\theta=0.055^{\circ}$ is the angle of the first minimum

$\lambda$ is the wavelength

Solving the equation for the wavelength, we find

$\lambda = 2 d sin \theta = 2(0.33\cdot 10^{-3} m)sin 0.055^{\circ}=6.33\cdot 10^{-7} m$