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In Young's double slit experiment the slit separation is 1 mm and the screen is 2 m away. The slits are illuminated by light containing wavelengths…

Numerical answer

In Young's double slit experiment the slit separation is

1\ \mathrm{mm}

and the screen is

2\ \mathrm{m}

away. The slits are illuminated by light containing wavelengths

600\ \mathrm{nm}

and

400\ \mathrm{nm}

. The smallest distance (in

\mu\mathrm{m}

) from the central maximum at which a bright fringe of the

600\ \mathrm{nm}

light coincides with a dark fringe of the

400\ \mathrm{nm}

light is

Show answer and solution

Answer: 1200

Position of the

n

th bright fringe of

600\ \mathrm{nm}

y=\dfrac{n\lambda_1D}{d}

. Dark fringe of

400\ \mathrm{nm}

y=\dfrac{(m+\frac12)\lambda_2D}{d}

. Equating,

600n=400\left(m+\tfrac12\right)

, i.e.

3n=2m+1

. The smallest solution with

n\ge1

n=1,\ m=1

. Then

y=\dfrac{1\times600\times10^{-9}\times2}{10^{-3}}=1.2\times10^{-3}\ \mathrm{m}=1200\ \mu\mathrm{m}

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