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 $\Delta Y=\beta$ )
$\theta_{0} =\frac{\beta}{D}=\frac{D \lambda}{d} \times \frac{1}{D}=\frac{\lambda}{d}$
$\theta_{0} =1^{\circ}=\pi / 180\, rad$
and $\lambda=6 \times 10^{-7} m$
$d =\frac{\lambda}{\theta_{0}}=\frac{180}{\pi} \times 6 \times 10^{-7}$
$=3.44 \times 10^{-5} m =0.03\, mm$