In Young's double slit experiment, the fifth maximum with wavelength ' $\lambda_1$ ' is at a distance '…
In Young's double slit experiment, the fifth maximum with wavelength ' $\lambda_1$ ' is at a distance ' $\mathrm{y}_1$ ' and the same maximum with wavelength ' $\lambda_2$ ' is at a distance ' $y_2$ ' measured from the central bright band. Then $\frac{y_1}{y_2}$ is equal to [D and $d$ are constant]
$\frac{\lambda_1}{\lambda_2}$
$\frac{\lambda_2}{\lambda_1}$
$\frac{\lambda_1^2}{\lambda_2^2}$
$\frac{\lambda_2^2}{\lambda_1^2}$
Solution
The equations for the position of the fringe from the central maxima are given as
$\begin{aligned}
\mathrm{y}_1 & =\frac{5 \lambda_1 \mathrm{D}}{\mathrm{d}} \\
\mathrm{y}_2 & =\frac{5 \lambda_2 \mathrm{D}}{\mathrm{d}} \\
\therefore \quad \frac{\mathrm{y}_1}{\mathrm{y}_2} & =\frac{\lambda_1}{\lambda_2}
\end{aligned}$
*