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In Young's double slit experiment, the $\mathrm{10^{\text {th }}}$ maximum of wavelength $\mathrm{\lambda_1}$ is at a distance of $\mathrm{y_1}$ from the central maximum. When the wavelength of the source is changed to $\mathrm{\lambda_2, 5^{\text {th }}}$ maximum is at a distance of $\mathrm{\mathrm{y}_2}$ from its central maximum. The ratio $\mathrm{\left(\frac{\mathrm{y}_1}{\mathrm{y}_2}\right)}$ is:
Option: 1
$\mathrm{\frac{2 \lambda_1}{\lambda_2}}$

Option: 2
$\mathrm{\frac{2 \lambda_2}{\lambda_1}}$

Option: 3
$\mathrm{\frac{\lambda_1}{2 \lambda_2}}$

Option: 4
$\mathrm{\frac{\lambda_2}{2 \lambda_1}}$

Answers (1)

best_answer

The distance of $10^{\text {th }}$ maximum of wavelength $\lambda_1$ from the central maximum is

$\mathrm{ \mathrm{y}_1=10 \lambda_1 \frac{\mathrm{D}}{\mathrm{d}} }$
Where $\mathrm{\mathrm{D}}$ is the distance of the slits from the screen and $\mathrm{d}$ is the distance between the slits. The distance of $5^{\text {th }}$ maximum of

wavelength $\lambda_2$ from the central maximum is

$\mathrm{ \begin{aligned} & \mathrm{y}_2=5 \lambda_2 \frac{\mathrm{D}}{\mathrm{d}} \\ & \therefore \quad \frac{\mathrm{y}_1}{\mathrm{y}_2}=\frac{2 \lambda_1}{\lambda_2} \end{aligned} }$

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