In an interference experiment, third bright ...

CBSE Class 12 Physics 2023 Delhi Set 3 Paper

Question : 6 of 14

Marks: +1, -0

In an interference experiment, third bright fringe is obtained at a point on the screen with a light of

700\text{ nm}

. What should be the wavelength of the light source in order to obtain the fifth bright fringe at the same point ?

Solution: 👈: Video Solution

Since,

y_n=\frac{n \lambda D}{d}

For

3^{\text{rd}}

bright fringe,

y_3=\frac{3 \times 700 D}{d}=\frac{2100 D}{d}

For

5^{\text{th}}

bright fringe,

y_5=5 \lambda' \frac{D}{d}

Since

y_5=y_3

5 \lambda' \frac{D}{d}=2100 \frac{D}{d}

∴

\lambda'=\frac{2100}{5}=420\text{ nm}

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