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Question: A parallel beam of light of wavelength 500nm falls on a narrow slit and the resulting diffraction pa...

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

A

0.2 nm

B

1 nm

C

2 nm

D

1.5 nm

Answer

0.2 nm

Explanation

Solution

Width of slit, d=nλDxd = \frac{n\lambda D}{x}

Here, n = 1, D = 1, x= 2.5mm=2.5×10−3m2.5mm = 2.5 \times 10^{- 3}m

λ=500nm=500×10−9m\lambda = 500nm = 500 \times 10^{- 9}m

∴d=1×500×10−9×12.5×10−3=2×10−4m=0.2mm\therefore d = \frac{1 \times 500 \times 10^{- 9} \times 1}{2.5 \times 10^{- 3}} = 2 \times 10^{- 4}m = 0.2mm