Question

Light with a wavelength of 612 nm incident on a double slit produces a second-order maximum at an angle of 25 degree. What is the separation between the slits?

Answer 1

Answer:

Speration between the slits $d=2.86\ \mu m$

Explanation:

Given that,

Wavelength of light, $\lambda=612\ nm=612\times 10^{-9}\ m$

It is incident on a double slit produces a second-order maximum at an angle of 25 degree.

The condition for maxima is given as :

$dsin\theta=m\lambda$

d = seperation between the slits

m = order, m = 2

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

$d=\dfrac{2\times 612\times 10^{-9}\ m}{sin(25)}$

d = 0.00000289

or

$d=2.86\mu m$

So, the seperation between the silts is $2.86\ \mu m$. Hence, this is the required solution.