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

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

Step-by-step explanation:

Given that,

Wavelength of light,
`\lambda=612\ nm=612* 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=(m\lambda)/(sin\theta)`

`d=(2* 612* 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.