Question

Consider light that has its third minimum at an angle of 23.3° when it falls on a single slit of width 4.05 μm. Find the wavelength of the light in nanometers.

Answer 1

Answer:

The wavelength of light is 533 nm.

Explanation:

It is given that,

Width of a single slit, $d=4.05\times 10^{-6}\ m$

Light has its third minimum at an angle of 23.3° when it falls on a single slit. For destructive interference, the equation for minima is given by:

$d\ sin\theta=n\lambda$  

Here, n = 3

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

$\lambda=\dfrac{4.05\times 10^{-6}\times sin(23.3)}{3}$

$\lambda=5.33\times 10^{-7}\ m$

$\lambda=533\ nm$

So, the wavelength of the light is 533 nm. Hence, this is the required solution.