Question

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

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

Answer 1

The correct option is A $5000\mathring{A}$

Path difference between ninth maxima and second minima, $7.5\times {10}^{-3}=\frac{\left(9\right)\lambda D}{d}-\frac{(2\times 2-1)\lambda D}{2d}=\frac{15\lambda D}{2d}$

$\lambda =\frac{7.5\times {10}^{-3}\times 2d}{15D}=\frac{7.5\times {10}^{-3}\times 2\times 0.5\times {10}^{-3}}{15\times 100\times {10}^{-2}}=0.5\times {10}^{-6}m=5000A^o$