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

Answer:

Distance between slits, $d=2.89\times 10^{-7}\ m$  

Explanation:

It is given that,

Wavelength, $\lambda=410\ nm=410\times 10^{-9}\ m$

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 :

$d\ sin\theta=(n+\dfrac{1}{2})\lambda$

For first minimum, n = 0

So, $d\ sin\theta=(\dfrac{1}{2})\lambda$

d is the distance between slits

So, $d=\dfrac{1/2\lambda}{sin\theta}$

$d=\dfrac{1/2\times 410\times 10^{-9}}{sin(45)}$

$d=2.89\times 10^{-7}\ m$

So, the distance between two slits is $2.89\times 10^{-7}\ m$. Hence, this is the required solution.