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. $3000 ?$
• B. $4000 ?$
• C. $6000 ?$
• D. $7000 ?$

Answer 1

Answer: (C)

$6000 ?$

Answer 2

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 $\lambda = \frac{(3\times 10^{-5} \,m)(10^{-3}\,m)}{(5\times 10^{-2})}$ $= 6 \times 10^{-7} \,m$ $= 6000 ?$