Question

In Fraunhofer diffraction pattern, slit width is 0.2 mm and screen is at 2 m away from the lens. If wavelength of light used is 5000 Å, then the distance between the first minimum on either side of the central maximum is ______. (θ is small and measured in radian)

Options

• 10^-1 m

• 10^-2 m

• 2 × 10^-2 m

• 2 × 10^-1 m

Answer 1

In Fraunhofer diffraction pattern, slit width is 0.2 mm and screen is at 2 m away from the lens. If wavelength of light used is 5000 Å, then the distance between the first minimum on either side of the central maximum is 10^-2 m. (θ is small and measured in radian)

Explanation:

Given d = 0.2 × 10^-3 m, D = 2m

and λ = 5 × 10^-7m

From B = `(lambda"D")/"d"=(5xx10^-7xx2)/(0.2xx10^-3)` 

= `(5xx10^-7)/10^-4`

∴ B = 510^-3 m

Distance between 1 st minima on either side

= 5 × 10^-3 m + 5 × 10^-3 

= 10 × 10^-3

= 10^-2 m