Question

In Young's double slit experiment, the spacing between the slits is $d$ and wavelength of light used is $6000\mathring{A}$. If the angular width of a fringe formed on a distance screen is $1^{\circ }$, then value of $d$ is

• A. 1 mm
• B. 0.05 mm
• C. 0.03 mm
• D. 0.01 mm

Answer 1

Answer: (C)

Here, $\sin \theta =\left(\frac{Y}{D}\right)$
So, $\Delta \theta =\frac{\Delta Y}{D}$
Angular fringe width ${\theta }_0=\Delta \theta$ (width $\Delta Y=\beta$ )
${\theta }_0=\frac{\beta }{D}=\frac{D\lambda }{d}\times \frac{1}{D}=\frac{\lambda }{d}$
${\theta }_0=1^{\circ }=\pi /180rad$
and $\lambda =6\times 10^{-7}m$
$d=\frac{\lambda }{{\theta }_0}=\frac{180}{\pi }\times 6\times 10^{-7}$
$=3.44\times 10^{-5}m=0.03mm$