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 $d\frac{6280\times 10^{-10}}{3.14}\times 180$
$3.6\times 10^{-5}m=0.036mm$