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.
Wavelength of light beam, λ = 500 nm = 500 × 10 -9 m
Distance of the screen from the slit, D = 1 m
For first minima, n = 1
Distance between the slits = d
Distance of the first minimum from the centre of the screen can be obtained as:
x = 2.5 mm = 2.5 × 10 -3 m
It is related to the order of minima as:
d = nλD/x = 1x500x10-9x1 / 2.5x10-3 = 2x10-4 = 0.2 mm
Therefore, the width of the slit is 0.2 mm.
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Welcome to the NCERT Solutions for Class 12 Physics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 19: A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction p....
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