CBSE Class 12 Physics 2023 Delhi Set 3 Paper

Question : 6

Total: 14

In an interference experiment, third bright fringe is obtained at a point on the screen with a light of 700‌nm. What should be the wavelength of the light source in order to obtain the fifth bright fringe at the same point ?

420‌nm
750‌nm
630‌nm
500‌nm

Solution: 👈: Video Solution

Since, yn=‌

nλD

For 3‌rd ‌ bright fringe,
y3=‌

3×700D

=‌

2100D

For 5‌th ‌ bright fringe,
y5=5λ′‌

Since y5=y3
5λ′‌

=2100‌

∴ λ′=‌

2100

=420‌nm

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