Optics 6 Question 54

56. Angular width of central maximum in the Fraunhofer diffraction pattern of a slit is measured. The slit is illuminated by light of wavelength 6000\AA. When the slit is illuminated by light of another wavelength, the angular width decreases by 30. Calculate the wavelength of this light. The same decrease in the angular width of central maximum is obtained when the original apparatus is immersed in a liquid. Find refractive index of the liquid.

(1996,2M)

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Answer:

Correct Answer: 56. 4200\AA,1.43

Solution:

  1. (a) Given, λ=6000\AA

Let b be the width of slit and D the distance between screen and slit.

First minima is obtained at bsinθ=λ

or bθ=λ as sinθθ or θ=λb

Angular width of first maxima =2θ=2λbλ

Angular width will decrease by 30 when λ is also decreased by 30.

Therefore, new wavelength

λ=(6000)301006000\AAλ=4200\AA

(b) When the apparatus is immersed in a liquid of refractive index μ, the wavelength is decreased μ times. Therefore,

4200\AA=6000\AAμμ=60004200 or μ=1.4291.43



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