Q101 / 120 · VITEEE 2009 · Physics
In Young's double slit experiment, the $10^{\text {th }}$ 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 $\lambda_{2}, 5^{\text {th }}$ maximum is at a distance of $y_{2}$ from its central maximum. The ratio $\left(\frac{\mathrm{y}_{1}}{\mathrm{y}_{2}}\right)$ is
- Correct answerA$\frac{2 \lambda_{1}}{\lambda_{2}}$
- B$\frac{2 \lambda_{2}}{\lambda_{1}}$
- C$\frac{\lambda_{1}}{2 \lambda_{2}}$
- D$\frac{\lambda_{2}}{2 \lambda_{1}}$
Solution
Position fringe from central maxima
$$
\begin{array}{l}
y_{1}=\frac{n \lambda_{1} D}{d} \\
\text { Given, } n=10 \\
\therefore y_{1}=\frac{10 \lambda_{1} D}{d}...(i)
\end{array}
$$
For second source
$$
\begin{array}{l}
y_{2}=\frac{5 \lambda_{2} D}{d}...(ii) \\
\therefore \frac{y_{1}}{y_{2}}=\frac{\frac{10 \lambda_{1} D}{d}}{\frac{5 \lambda_{2} D}{d}}=\frac{2 \lambda_{1}}{\lambda_{2}}
\end{array}
$$
Hence, the correct answer is option: 'A'.