**Answer the following question.**

In the diffraction due to a single slit experiment, the aperture of the slit is 3 mm. If monochromatic light of wavelength 620 nm is incident normally on the slit, calculate the separation between the first order minima and the 3rd order maxima on one side of the screen. The distance between the slit and the screen is 1.5 m.

#### Solution

As we know that the angular position of the minimum intensity on the screen is given as

`asintheta = Nlambda`

so we have for first-order minimum

`(3 xx 10^-3)sin theta = 620 xx 10^-9`

`theta_1 = 0.0118^circ`

now similarly for 3rd order minimum

`atheta_2 = 3lambda`

`(3 xx 10^-3)sintheta_2 = 3(620 xx 10^-9)`

`theta_2 = 0.0355^circ`

now the angular seperation between two minimum

`Deltatheta = 0.0355 - 0.0118`

`Deltatheta = 0.0237^circ`

So the distance between two minimum is

`d = LDeltatheta`

`d = 1.5(0.0237 xx pi/180)`

d = 0.62 mm.