Question

In a double slit experiment, the distance between the two slits is $0.6\text{mm}$ and these are illuminated with light of wavelength $4800\mathring{A}$. The angular width of dark fringe on the screen at a distance $120\text{cm}$ from slits will be,

• A. $8\times {10}^{-4}\text{rad}$
• B. $6\times {10}^{-4}\text{rad}$
• C. $4\times {10}^{-4}\text{rad}$
• D. $16\times {10}^{-4}\text{rad}$

Answer 1

The correct option is C $4\times {10}^{-4}\text{rad}$

Given :-
$d=0.6\text{mm}=0.6\times {10}^{-3}\text{m}$
$\lambda =4800\mathring{A}=4800\times {10}^{-10}\text{m}$
$D=120\text{cm}=1.2\text{m}$

Angular width of dark fringe is equal to angular fringe width.
So, $\theta =\frac{\lambda }{d}=\frac{4800\times {10}^{-10}}{0.6\times {10}^{-3}}=8\times {10}^{-4}\text{rad}$

$$\begin{array}{c}\text{Why this question :}\\ \text{Caution : Be cautious if question is asking for angular width of dark fringe or angular position of dark fringe. Angular width is the angle between successive dark fringes. If angular position of first minima was asked then,}\theta =\frac{\lambda }{2d}.\end{array}$$