Question

In a Young’s double slit experiment the angular width of a fringe formed on a distant screen is $1^{\circ}$. The wavelength fo the light used is $6280 \,\mathring{A}$. What is the distance between the two coherent sources?

• A. 0.036 mm
• B. 0.12 mm
• C. 6 mm
• D. 4 mm

Answer 1

Answer: (A)

The angular fringe width is given by $\alpha = \frac{\lambda}{d} $
where $l$ is wavelength and $\lambda $ is the distance between two coherent sources. Thus
$d = \frac {\lambda}{\alpha}$
Given, $\lambda = 6280 \,\mathring{A}, \alpha 1^{\circ} = \frac {\pi}{180}$ radian.
Thus $d \frac {6280 \times 10^{-10}}{3.14} \times 180$
$3.6 \times 10^{-5} m = 0.036\, mm $