Question

In Young's double slit experiment the slits are 0.5 mm apart and the interference is observed on a screen at a distance of 100 cm from the slit. It is found that the 9th bright fringe is at a distance of 7.5 mm from the second dark fringe from the centre of the fringe pattern. The wavelength of the light used is

• A. $\frac{2500}{7}{A}^{\circ }$
• B. $2500A^{\circ }$
• C. $5000A^{\circ }$
• D. $\frac{5000}{7}{A}^{\circ }$

Answer 1

The correct option is C $5000A^{\circ }$

Given,

$d=0.5mm$

$D=100cm$

$X_9-X_2^{\prime }=7.5mm$. . . . . . . .(1)

Condition for bright fringe,

$X_n=\frac{n\lambda D}{d}$

for $9^{th}$ bright fringe

$X_9=\frac{9\lambda D}{d}$

Condition for dark fringe,

$X_n=\frac{(2n-1)}{2d}$

$X_2^{\prime }=\frac{(2\times 2-1)\lambda D}{2d}=\frac{3\lambda D}{2d}$

From equation (1)

$\frac{9\lambda D}{d}-\frac{3\lambda D}{2d}=7.5\times {10}^{-3}m$

$\frac{15\lambda D}{2d}=7.5mm$

$\lambda =\frac{2d}{15D}\times 7.5mm$

$\lambda =\frac{2\times 0.5mm}{15\times 100cm}\times 7.5mm$

$\lambda =5000\times {10}^{-10}m=5000A^{\circ }$

The correct option is C.