Question

In Young's double slit experiment, while using a source of light of wavelength 5000Å, the fringe width obtained is 0.6 cm. If the distance between the screen and slit is reduced to half, the wavelength of the source to get fringes 0.003 m wide is :

• A. $4\times {10}^{-7}$m
• B. $5\times {10}^{-7}$m
• C. $10\times {10}^{-7}$m
• D. $2.5\times {10}^{-7}$m

Answer 1

The correct option is B $5\times {10}^{-7}$m

$\beta =\frac{D\lambda }{d}$
${\beta }_1=\frac{D_1{\lambda }_1}{d}$
$0.6\times {10}^{-2}=\frac{D_{1}5000\times {10}^{-10}}{d}\Rightarrow \frac{d}{D_1}=\frac{5000\times {10}^{-10}}{0.6\times {10}^{-2}}$
${\beta }_2=0.003m$
${\beta }_2=\frac{D_2{\lambda }_2}{d}$
$0.003=\frac{D_2{\lambda }_2}{2d}(D_2=\frac{1}{2}{D}_1$ as given in problem)
$0.003=\frac{D_1{\lambda }_2}{2d}$
${\lambda }_2=\frac{0.003\times 2\times d}{D_1}$
$=0.003\times 2\times \frac{5000\times {10}^{-10}}{0.6\times {10}^{-2}}\left(From1\right)$
$=50\times {10}^{-8}$
$=5\times {10}^{-7}m$