Question

In Young's double slit experiment, the light emitted from source has $\lambda =6.5\times 10^{-7}m$ and the distance between the two slits is 1 mm. Distance between the screen and slit is 1 metre. Distance between third dark and fifth bright fringe will be

• A. 3.2 mm
• B. 1.63 mm
• C. 0.585 mm
• D. 2.31 mm

Answer 1

Answer: (B)

$x_5=n\frac{\lambda D}{d}=\frac{5\times 6.5\times 10^{-7}\times 1}{10^{-3}}=32.5\times 10^{-4}m$
${\mathrm{x}}_3=(2\mathrm{n}-1)\frac{\lambda \mathrm{D}}{2\mathrm{d}}=\frac{5\times 6.3\times 10^{-7}\times 1}{2\times 10^{-3}}=16.25\times 10^{-4}\mathrm{m}$
$\therefore x_5-x_3\approx 1.63mm$