Class 12 Physics
CBSE and UP Board
Wave Optics
Chapter-10 Exercise 10.19
NCERT Solutions for Class 12 Physics Chapter 10 Question-19
Additional Exercise
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.
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Wavelength of light beam, λ= 500 nm = 500 x 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 x 10-3 m
It is related to the order of minima as: d
nλ = x d/D
d= nλ D/x
= 1 x 500 x 10⁻9 x 1/(2.5 x 10-3 ) = 2 x 10-4 m= 0.2 mm
Therefore, the width of the slits is 0.2 mm