CBSE Class 12 Physics 2022 Term 2 Outside Delhi Set 3 Solved Paper

Question : 3 of 5

Marks: +1, -0

A beam of light consisting of two wavelengths

600\text{ nm}

and

500\text{ nm}

is used in Young's double slit experiment. The silt separation is

1.0\text{ mm}

and the screen is kept

0.60\text{ m}

away from the plane of the slits. Calculate:

(i) the distance of the second bright fringe from the central maximum for wavelength

500\text{ nm}

, and

(ii) the least distance from the central maximum where the bright fringes due to both the wavelengths coincide.

Solution:

(i) Distance of 2nd bright fringe from central maximum

=\frac{2\lambda D}{d}

=\frac{2\times500\times10^{-9}\times0.6}{1\times10^{-3}}

=6\times10^{-4}\text{ m}

(ii)

\frac{n\lambda_1 D}{d} = \frac{(n+1)\lambda_2 D}{d}

Or,

n\lambda_1 = (n+1)\lambda_2

Or,

\frac{n}{n+1} = \frac{\lambda_2}{\lambda_1}

Or,

\frac{n}{n+1} = \frac{500}{600}

\therefore n=5

So, least distance from central maximum

=\frac{5\times600\times10^{-9}\times0.6}{1\times10^{-3}} = 18\times10^{-4}\text{ m}

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