Question

If in a Young's double slit experiment, the slit width is $3 \,cm$, the separation between slits and screen is $7\, cm$ and wavelength of light is $1000 \,\mathring{A}$, then fringe width will be $\left(\alpha=1^{\circ}, \mu=1.5\right)$ :

• A. $2 \times 10^{-5} \,m$
• B. $2 \times 10^{-3}\, m$
• C. $0.2 \times 10^{-5}\, m$
• D. none of these

Answer 1

Answer: (D)

Fringe width
$\beta=\frac{\lambda D}{d}\,\,\,...(1)$
Given : $\lambda=1000\,\mathring{A}=10^{-5} \,cm , $
$d=3\, cm$,
$D=7 \,cm$
Putting given value in eq (i)
$\therefore \beta =\frac{10^{-5} \times 7}{3} $
$=2.3 \times 10^{-5} \,cm$
$ =2.3 \times 10^{-7}\, m$