JEE Main 2021 — Wave Optics Question with Solution

In the double-slit experiment, the position of a bright fringe is given by the formula :

$y=\frac{m\lambda L}{d}$,

where :

• m is the order of the fringe (1 for the first bright fringe, 2 for the second, etc.)
• $\lambda$ is the wavelength of the light (in meters)
• L is the distance from the slits to the screen (in meters)
• d is the distance between the slits (in meters)

We need to find the difference in position between the first and third bright fringes. So, we find the position of both and subtract the position of the first from the position of the third :

$y_3=\frac{3\lambda L}{d}$

$y_1=\frac{\lambda L}{d}$

Then, the difference between the third and the first bright fringes is :

$y_3-y_1=2\frac{\lambda L}{d}$

Now, let's plug the given values: $\lambda =5890\mathring{A}=5890\times 10^{-10}m$ (since 1 Å = $10^{-10}$ meters), L = 0.5 m, and d = 0.5 mm = 0.5 $\times 10^{-3}$ m :

$$y_3-y_1=2\times \frac{5890\times 10^{-10}\text{m}\times 0.5\text{m}}{0.5\times 10^{-3}\text{m}}=1178\times 10^{-6}\text{m}$$

So, the correct answer is 1178 $\times 10^{-6}$ m, which corresponds to Option B.