Question

The third-order bright fringe of 610-nm light is observed at an angle of 31° when the light falls on two narrow slits. How far apart are the slits?

Answer 1

Answer:

The distance between the slits is 3.55 μm

Explanation:

Given that,

Order number = 3

Wave length = 610 nm

Angle = 31°

We need to calculate the distance between the slits

Using formula of distance of slit

$d\sin\theta=n\lambda$

$d=\dfrac{n\lambda}{\sin\theta}$

Where, n = order number

$\lambda$= wavelength

Put the value into the formula

$d=\dfrac{3\times610\times10^{-9}}{\sin31}$

$d=3.55\times10^{-6}\ m$

$d= 3.55\ \mu m$

Hence, The distance between the slits is 3.55 μm.