Question

In Young's double slit experiment, the fringe width is $1\times {10}^{-4}m$, if the distance between the slit and screen is doubled and the distance between two slits is reduced to half and the wavelength is changed from $6.4\times {10}^{-7}m$ to $4.0\times {10}^{-7}m$, then the value of new fringe width will be

• A. $2.5\times {10}^{-4}m$
• B. $2.0\times {10}^{-4}m$
• C. $1.25\times {10}^{-4}m$
• D. $0.15\times {10}^{-4}m$

Answer 1

The correct option is A $2.5\times {10}^{-4}m$

Fringe width, $\beta =\frac{D\lambda }{d}$
According to question, $\frac{{\beta }_2}{{\beta }_1}=\frac{{\lambda }_2{D}_2{d}_1}{{\lambda }_1{D}_1{d}_2}$
Here, $D_2=2D_1$
$d_2=\frac{d_1}{2},\frac{{\beta }_2}{{\beta }_1}=\frac{{\lambda }_2\left(2D_1\right)\times d_1}{{\lambda }_1\left(D_1\right)\times \left(\frac{d_1}{2}\right)}$
$=4\frac{{\lambda }_2}{{\lambda }_1}=\frac{4\times 4\times {10}^{-7}}{6.4\times {10}^{-7}}$
$\beta =\frac{4\times 4\times {10}^{-7}}{6.4\times {10}^{-7}}=2.5\times {10}^{-4}m$.