Question

In Young's double slit experiment, the distance between the slits is $1mm$ and screen is $25cm$ away from the slits. If the wavelength of light is $6000\mathring{A}$, the fringe width on the screen is:

• A. $0.15mm$
• B. $0.30mm$
• C. $0.24mm$
• D. $0.12mm$

Answer 1

The correct option is A $0.15mm$

All bright and dark fringes are of equal width.
Fringe width is given by
$\beta =\frac{D\lambda }{d}$
where $d$ is separation between slits,
$D$ is distance between slit and screen,
$\lambda$ is wavelength of monochromatic light.
Here, $\lambda =6000\mathring{A}=6000\times {10}^{-10}m$
$D=25cm=25\times {10}^{-2}m$
$d=1mm=1\times {10}^{-3}$
Hence, $\beta =\frac{6000\times {10}^{-10}\times 25\times {10}^{-2}}{1\times {10}^{-3}}$
$=1.5\times {10}^{-4}$
$=0.15\times {10}^{-3}m=0.15mm$