Physics Class 12 NCERT Solutions: Chapter 10 Wave Optics Part 2

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Young's double slits experiment

Young's Double Slits Experiment

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Q: 4. In a Young’s double-slit experiment, the slits are separated by and the screen is placed away. The distance between the central bright fringe and the fourth bright fringe is measured to be . Determine the wavelength of light used in the experiment.


Distance between the slits,

Distance between the slits and the screen,

Distance between the central fringe and the fourth fringe,

In case of a constructive interference, we have the relation for the distance between the two fringes as:


Order of fringes

Wavelength of light used

Hence, the wavelength of the light is.

Q: 5. In Young’s double-slit experiment using monochromatic light of wavelength , the intensity of light at a point on the screen where path difference is, is units. What is the intensity of light at a point where path difference is ?


Let be the intensity of the two light waves. Their resultant intensities can be obtained as:


Phase difference between the two waves

For monochromatic light waves,

Phase difference path difference

Since path difference,

Phase difference,


When path difference,

Phase difference,

Hence, resultant intensity,

Using equation (1), we can write:

Hence, the intensity of light at a point where the path difference is is units.

Q: 6. A beam of light consisting of two wavelengths, and, is used to obtain interference fringes in a Young’s double-slit experiment.

(A) Find the distance of the third bright fringe on the screen from the central maximum for wavelength.

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


Wavelength of the light beam,

Wavelength of another light beam,

Distance of the slits from the screen = D

Distance between the two slits = d

(A) Distance of the nth bright fringe on the screen from the central maximum is given by the relation,

For third bright fringe.

(B) Let the nth bright fringe due to wavelength and bright fringe due to wavelength coincide on the screen. We can equate the conditions for bright fringes as:

Hence, the least distance from the central maximum can be obtained by the relation:

Note: The value of d and are not given in the question.