wave-optics Question 17

Question: Q. 6. A beam of light consisting of two wavelengths 800 nm and 600 nm is used to obtain the interference fringes in a Young’s double slit experiment on a screen placed 1.4 m away. If the two slits are separated by 0.28 mm, calculate the least distance from the central bright maxima where the bright fringes of the two wavelengths coincide.

A [O.D. I, II, III 2012]

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Solution:

Ans. The two bright fringes will coincide when

mλ1=(m+1)λ2

m×800×109=(m+1)×600×109

m=3

Xm=mDλ1d

Xm=3×1.4×800×1090.28×103 m

Xm=12×108 m

1/2

[CBSE Marking Scheme 2012]

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Q7. Name the phenomenon which is responsible for bending of light around sharp corners of an obstacle. Under what conditions does this phenomenon take place? Give one application of this phenomenon in everyday life.

R [SQP 2014]

Ans. Diffraction.

Diffraction Condition : The size of the obstacle sharpness should be comparable to the wavelength of the light falling. 1+1/2 Application : The finite resolution of our eye. 1/2

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[AI Q. 8. Write the distinguishing features between a diffraction pattern due to a single slit and the interference fringes produced in Young’s double slit experiment. [OD Comptt. I, II, III 2013]

Ans. Diffraction due to a Single Slit :

(i) It is produced due to different parts of same wavefront.

(ii) Central fringe is twice as wide as other fringes.

(iii) Intensity of fringes decreases as we go to successive maxima away from the centre.

(iv) At an angle λa, first minima is obtained.

Interference Fringe due to Young’s Double Slit :

(i) It is produced due to two different wavefronts.

(ii) Fringe width is of same size.

(iii) Fringes have same intensity.

(iv) At an angle λa, maxima is obtained.



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