In a Young’s double-slit experimen | Class 12 Physics Chapter Wave Optics, Wave Optics NCERT Solutions

Question:

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.

Answer:

Here it is given that,

Distance between the slits, d = 0.28 mm = 0.28 × 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 × 10 -2 m

For constructive interference, the distance between the two fringes is given by relation: u = nλ D/d

where, n = Order of fringes

wavelength of the light can be given as: λ = ud/nD = 1.2x10-2x0.28x10-3/4x1.4 = 6x10-7 = 600 nm

Hence, the wavelength of the light is 6 x 10 -7 m.


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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 4: In a Young’s double-slit experiment, the slits are separated by 0.28 mm and the screen is plac....