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

The wavelength of light is 533 nm.

Step-by-step explanation:

It is given that,

Width of a single slit,
`d=4.05* 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=(d\ sin\theta)/(n)`

`\lambda=(4.05* 10^(-6)* sin(23.3))/(3)`

`\lambda=5.33* 10^(-7)\ m`

`\lambda=533\ nm`

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