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$
$= 6000 \mathring{A}$