Optics 6 Question 57

59. A beam of light consisting of two wavelengths, 6500\AA and 5200\AA is used to obtain interference fringe in a Young’s double slit experiment.

(1985,6 M)

(a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 6500\AA.

(b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?

The distance between the slits is 2mm and the distance between the plane of the slits and the screen is 120cm.

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Solution:

  1. (a) The desired distance will be

y1=3ω1=3λ1Dd=(3)(6500×1010)(1.2)(2×103)=11.7×104m=1.17mm

(b) Let n1 bright fringe of λ1 coincides with n2 bright fringe of λ2. Then,

n1λ1Dd=n2λ2Dd or n1n2=λ2λ1=52006500=45

Therefore, 4th bright of λ1 coincides with 5th bright of λ2. Similarly, 8th bright of λ1 will coincide with 10th bright of λ2 and so on. The least distance from the central maximum will therefore corresponding to 4 th bright of λ1 (or 5 th bright of λ2.) Hence,

Ymin =4λ1Dd=4(6500×1010)(1.2)(2×103)=15.6×104m=1.56mm



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