Question

In a two slit experiment with monochromatic light, fringes are obtained on a screen placed at some distance from the plane of slits. If the screen is moved by $5\times 10^{-2}m$ towards the slits, the change in fringe width is $3\times 10^{-5}m$. If the distance between slits is $10^{-3}m$, the wavelength of light will be

• A. $3000\mathring{A}$
• B. $4000\mathring{A}$
• C. $6000\mathring{A}$
• D. $7000\mathring{A}$

Answer 1

Answer: (C)

Fringe width, $\beta =\frac{\lambda D}{d}$
where $\lambda$ is the wavelength of light, $D$ is the distance between screen and the slits and $d$ is the distance between two slits
$\therefore \Delta \beta =\frac{\lambda }{d}\Delta D$ or
$\lambda =\frac{\Delta \beta d}{\Delta D}$ ( As $\lambda$ and $d$ are constants)
Substituting the given values, we get
$\lambda =\frac{(3\times 10^{-5}m)(10^{-3}m)}{(5\times 10^{-2})}$
$=6\times 10^{-7}m$
$=6000\mathring{A}$