Question

In Young's double slit experiment, fringes of width $\beta$ are produced on a screen kept at a distance of $1m$ from the slit. When the screen is moved away by $5\times {10}^{-2}m$, fringe width changes by $3\times {10}^{-5}m$. The separation between the slits is $1\times {10}^{-3}m$. The wavelength of the light used is ________ $nm$.

• A. $500$
• B. $600$
• C. $700$
• D. $400$

Answer 1

The correct option is B $600$

Fringe width $\beta =\frac{\lambda D}{d}$

Thus change in fringe width $\Delta \beta =\frac{\lambda }{d}\Delta D$

Given : $\Delta D=5\times {10}^{-2}m$, $d=1\times {10}^{-3}m$ , $\Delta \beta =3\times {10}^{-5}m$

$\therefore$ $3\times {10}^{-5}=\frac{\lambda }{1\times {10}^{-3}}\times 5\times {10}^{-2}$

$⟹$ $\lambda =6\times {10}^{-7}$ $m=600$ $nm$