Question 292: Wave Optics

Question

In young's double slit experiment, the $\mathrm{n}^{	ext {th }}$ 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 $\lambda_2,\left(\frac{\mathrm{n}}{3}ight)^{	ext {th }}$ maximum is at a distance of $y_2$ from its central maximum. The ratio $\frac{y_1}{y_2}$ is

Accepted Answer

In Young's double-slit experiment, the position of the nth maximum for a wavelength $\lambda_1$ is given by:
$  y_1 = \frac{n \lambda_1 D}{d} \quad 	ext{(Equation 1)} $
where $ D $ is the distance from the slits to the screen, and $ d $ is the separation between the slits.
When the wavelength is changed to $\lambda_2$, the position of the $\left(\frac{n}{3}ight)^{	ext{th}}$ maximum is given by:
$  y_2 = \frac{\frac{n}{3} \lambda_2 D}{d} \quad 	ext{(Equation 2)} $
To find the ratio $\frac{y_1}{y_2}$, substitute the expressions for $y_1$ and $y_2$ from Equations 1 and 2:
$ \frac{y_1}{y_2} = \frac{\frac{n \lambda_1 D}{d}}{\frac{\frac{n}{3} \lambda_2 D}{d}} $
Simplifying this expression:
$ \frac{y_1}{y_2} = \frac{3 \lambda_1}{\lambda_2} $
Thus, the ratio $\frac{y_1}{y_2}$ is $\frac{3 \lambda_1}{\lambda_2}$.

Answer

$\frac{3 \lambda_1}{\lambda_2}$

Answer

$\frac{3 \lambda_2}{\lambda_1}$

Answer

$\frac{\lambda_1}{3 \lambda_2}$

Answer

$\frac{\lambda_2}{3 \lambda_1}$

Wave Optics

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