Class 12 Physics
CBSE and UP Board
Wave Optics
Chapter-10 Exercise 10.4
NCERT Solutions for Class 12 Physics Chapter 10 Question-4
In a Young’s double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.
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Distance between the slits, d = 0.28 mm = 0.28 x 10-3 m
Distance between the slits and the screen, D = 1.4 m
Distance between the central fringe and the fourth (n = 4) fringe, u = 1.2 cm = 1.2 x 10-2 m
In case of a constructive interference, we have the relation for the distance between the two fringes as: u = n λ D/d
Where, n = Order of fringes = 4 λ = Wavelength of light used.
Therefore ,λ = ud/(nD)
= (1.2 x 10-2 x 0.28 x 10-3)/(4 x 1.4)
4×1.4
= 6 x 10–7 = 600 nm
Hence, the wavelength of the light is 600 nm.