Question

In Young's double slit experiment, fringes of width $\beta$ are produced on the 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 light used is __________$nm$.

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

Answer 1

The correct option is B $600$

Since, we know that in Young's double slit experiment.

$\beta =\frac{D\lambda }{d}$ ...(i)

${\beta }^{{}^{\prime }}=\frac{D^{{}^{\prime }}\lambda }{d}$ ...(ii)

Subtract equation (ii) from equation (i), we get

$\beta -{\beta }^{{}^{\prime }}=\frac{\lambda (D-D^{{}^{\prime }})}{d}$

Substituting the given values, we get

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

$\Rightarrow \lambda =\frac{3\times {10}^{-8}}{5\times {10}^{-2}}=0.6\times {10}^{-6}m$

$\Rightarrow \lambda =600\times {10}^{-9}m$

$\Rightarrow \lambda =600nm$