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NCERT Class XII Chapter
Wave Optics
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Question : 19 of 21
Marks: +1, -0
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
Solution:  
First minimum is observed at a distance 2.5 mm from centre of the screen.
So, x = 2.5 mm = 2.5 × 10−310^{-3} m
l = 500 nm = 500 × 10−910^{-9} m = 5 × 10−710^{-7} m
D = 1 m, n = 1
So, condition for minima is dxD\frac{dx}{D} = nλ
∴ Slit width = nλDx\frac{n\lambda D}{x}
∴ d = 5×10−7×1×12.5×10−3\frac{5 \times 10^{-7} \times 1 \times 1}{2.5 \times 10^{-3}} = 0.2 mm
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