Question

In Young's double slit experiment, the ${10}^{th}$ maximum of wavelength ${\lambda }_1$ is at a distance of $y_1$ from the central maximum. When the wavelength of the source is changed to ${\lambda }_2,5^{th}$ maximum is at a distance of $y_2$ from its central maximum. The ratio $\left(\frac{y_1}{y_2}\right)$ is

• A. $\frac{2{\lambda }_2}{{\lambda }_1}$
• B. $\frac{{\lambda }_1}{{2\lambda }_2}$
• C. $\frac{{\lambda }_2}{{2\lambda }_1}$
• D. $\frac{{2\lambda }_1}{{\lambda }_2}$

Answer 1

The correct option is D $\frac{{2\lambda }_1}{{\lambda }_2}$

$\begin{array}{c}\text{Position of nth maximum}=n\frac{\lambda D}{d}\\ \text{where,}\\ \lambda \text{is wave length of source}\\ \text{D is distance between sources and screen}\\ \text{d is distance between slits.}\end{array}$

$\text{In case:}1\lambda ={\lambda }_1\text{and}n\text{=}10$

$\therefore y_1=10\frac{{\lambda }_{1}D}{d}$

$\text{In case:}2,\lambda \text{=}{\lambda }_2\text{and n =}5$

$\begin{array}{c}\therefore y_2=\frac{5{\lambda }_{2}D}{d}\\ Hence,\frac{y_1}{y_2}=\frac{2{\lambda }_1}{{\lambda }_2}\end{array}$