Question

The fringe width in a Young's double slit interference pattern is $2.4\times {10}^{-4}m$ , when red light of wavelength $6400$ $\dot{A}$ is used. How much will it change, if blue light of wavelength $4000$$\dot{A}$ is used?

• A. $0.9\times {10}^{-4}m$
• B. $4.5\times {10}^{-4}m$
• C. $9\times {10}^{-4}m$
• D. $0.45\times {10}^{-4}m$

Answer 1

The correct option is A $0.9\times {10}^{-4}m$

$\begin{array}{c}\beta =\lambda \frac{D}{d}\\ 2.4\times {10}^{-4}=6400\times {10}^{-10}\times \frac{D}{d}\\ \frac{D}{d}=\frac{2.4\times {10}^{-4}}{6400\times {10}^{-10}}\\ {\beta }^{\prime }=4000\times {10}^{-10}\times \frac{2.4\times {10}^{-4}}{6400\times {10}^{-10}}\\ =1.5\times {10}^{-4}\\ \beta -{\beta }^{\prime }=\left(2.4-1.5\right){10}^{-4}\\ =0.9\times {10}^{-4}m\end{array}$

Hence, Option $A$ is correct.