Question

In Youngs double-slit experiment, the angular width of fringe formed on a distant screen is $ 0.1{}^\circ $ . If wavelength of light used is $ 6000\,\mathring{A} $ then spacing between the slits is:

• A. $ 3.44\times {{10}^{-4}}m $
• B. $ 4.33\times {{10}^{-4}}m $
• C. $ 5.44\times {{10}^{-4}}m $
• D. $ 6.33\times {{10}^{-4}}m $

Answer 1

Answer: (A)

Angular fringe width $\alpha=0.1^{\circ}$
$=\frac{0.1 \times \pi}{180}$
$=1.74 \times 10^{-3} rad$
$\lambda =6000 \,\mathring{A} $
$=6000 \times 10^{-10} m $
$=6 \times 10^{-7} m$
The angular fringe width is
$\alpha-\frac{\lambda}{d}$
(where $d$ is the separation between the slits)
$d =\frac{\lambda}{\alpha} $
$=\frac{6 \times 10^{-7}}{1.746 \times 10^{-3}}$
$=3.44 \times 10^{-4} m $