Q10.19 A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1 m away. It is observed that the first minimum is at a distance of 2.5 mm from the centre of the screen. Find the width of the slit.

Answers (1)

Given

Distance of screen from the slit, D=1m

Distance of first minimum  X_1=2.5mm=10^{-3}=2.5*10^{-3}mm

The wavelength of the light \lambda=500nm=500*10^{-9}m

Now,

As we know,

X_n=n\frac{\lambda D}{d}

d=n\frac{\lambda D}{X_n}=1*\frac{500*10^{-9}*1}{2.5*10^{-3}}=2*10^{-4}m=0.2mm

Hence, the width of the slit is 0.2 mm. 

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