Question 240: Waves Optics

Question

In Young's double slit experiment, the distance between the slits is 2 mm and the slits are 1 m away, from the screen. Two interference patterns can be obtained on the screen due to light of wavelength ' $\lambda_1$ ' and ' $\lambda_2$ ' respectively. The separation on the screen between the $3^{	ext {rd }}$ order bright fringes on the two interference patterns is ( $\lambda_2=1.5 \lambda_1$ )

Accepted Answer

The fringe position of $\mathrm{n}^{	ext {th }}$ order bright fringe is given by
$$ y=\frac{n \lambda D}{d} $$
For wavelength $\lambda_1$
$$ y_1=\frac{3 \lambda_1 D}{d} $$
For wavelength $\lambda_2=1.5 \lambda_1$
$$ y_2=\frac{3\left(1.5 \lambda_1ight) D}{d}=\frac{4.5 \lambda_1 D}{d} $$
So, the separation $\Delta y$ between the two $3^{	ext {rd }}$ order bright fringes is:
$$ \begin{aligned} & \Delta y=y_2-y_1=\frac{4.5 \lambda_1 D}{d}-\frac{3 \lambda_1 D}{d}=\frac{1.5 \lambda_1 D}{d} \ 	herefore \quad & \Delta y=\frac{1.5 \lambda_1 \cdot 1}{2 	imes 10^{-3}}=\frac{1.5 \lambda_1}{2 	imes 10^{-3}}=\left(0.75 	imes 10^3ight) \lambda_1 \end{aligned} $$

Answer

$\left(0.75 	imes 10^{+3}ight) \lambda_1$

Answer

$\left(1.75 	imes 10^{+3}ight) \lambda_1$

Answer

$\left(2.00 	imes 10^{+3}ight) \lambda_1$

Answer

$\left(2.25 	imes 10^{+3}ight) \lambda_1$

Waves Optics

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