Question

In Youngs double slit experiment, the spacing between the slits is $d$ and wave length of light used is $6000\,?$ . If the angular width of a fringe formed on a distant screen is $1{}^\circ ,$ then value of $d$ is:

• A. 1 mm
• B. 0.0.5 mm
• C. 0.03 mm
• D. 0.01 mm

Answer 1

Answer: (C)

0.03 mm

Answer 2

$\sin \theta \simeq \theta=\frac{y}{D}$ So, $\Delta \theta=\frac{\Delta y}{D}$. Angular fringe width $\theta_{0}=\Delta \theta$ (width $\Delta y=\beta)$ $\theta_{0}=\frac{\beta}{D}=\frac{D \lambda}{d} \times \frac{1}{D}=\frac{\lambda}{d}$ $\theta_{0}=1^{\circ}=\frac{\pi}{180} rad$ And $\lambda=6 \times 10^{-7} m$ $d =\frac{\lambda}{\theta_{0}}=\frac{180}{\pi} \times 6 \times 10^{-7}$ $=3.44 \times 10^{-5} m$ $=0.03\, mm$