Question

In a Young's double slit experiment, the separation between the two slits is $\frac{2}{\pi }mm$. The distance of the screen from the slits is $50cm$. If the wavelength of light used is $4000A^{\circ }$, then the angular position of first dark fringe is

• A. ${1.21}^{\circ }$
• B. ${0.018}^{\circ }$
• C. ${0.16}^{\circ }$
• D. ${0.3}^{\circ }$

Answer 1

The correct option is B ${0.018}^{\circ }$

Given
$D=50cm$
$d=\frac{2}{\pi }mm$
$\lambda =4000A^{\circ }$

From figure
$tan\theta =\frac{\beta }{2D}=\frac{\lambda D}{2dD}=\frac{\lambda }{2d}$

$\theta$ is small, $tan\theta =\theta$
$\theta =\frac{\lambda }{2d}$
$=\frac{4\times {10}^{-7}}{2\times \frac{2}{\pi }\times {10}^{-3}}$
$=\pi \times {10}^{-4}rad$
$=\pi \times {10}^{-4}\times \frac{180}{\pi }deg$
$={0.018}^o$