Question

Find the distance between two slits that produces the first minimum for 410-nm violet light at an angle of 45.0°

Answer 1

Distance between slits,
`d=2.89* 10^(-7)\ m`

Step-by-step explanation:

It is given that,

Wavelength,
`\lambda=410\ nm=410* 10^(-9)\ m`

Angle,
`\theta=45`

We need to find the distance between two slits that produces first minimum. The equation for the destructive interference is given by :

`d\ sin\theta=(n+(1)/(2))\lambda`

For first minimum, n = 0

So,
`d\ sin\theta=((1)/(2))\lambda`

d is the distance between slits

So,
`d=(1/2\lambda)/(sin\theta)`

`d=(1/2* 410* 10^(-9))/(sin(45))`

`d=2.89* 10^(-7)\ m`

So, the distance between two slits is
`2.89* 10^(-7)\ m`. Hence, this is the required solution.