Question

In Young's double slit experiment using $\lambda =6000A^0,$ distance between the screen and the source is 1m. If the fringe width on the screen is 0.06 cm, the distance between the two coherent sources is

• A. 0.01 mm
• B. 1 cm
• C. 0.1 mm
• D. 1 mm

Answer 1

The correct option is D 1 mm

$\beta =\left(\frac{\lambda D}{d}\right)$
$\Rightarrow d=\left(\frac{\lambda D}{\beta }\right)=\frac{6000\times {10}^{-10}\times 1}{0.06\times {10}^{-2}}$
$=\frac{6000\times {10}^{-8}\times 100}{0.06}$
$=\left({10}^{-3}\right)m$
$=1mm$