QJEEM23Ph32a2: JEE Main 2023 Physics - Class 12 JEEM23Ph32a2 Solution: In Young's double slits experiment, the position of 5 bright…

In Young's double-slit experiment, the distance between the slits and the screen, $L = 1 \text{ m}$ , the wavelength of the light, $\lambda = 600 \text{ nm} = 600 \times 10^{-9} \text{ m}$ , and the position of the 5th bright fringe from the central maximum, $y = 5 \text{ cm} = 0.05 \text{ m}$ .

The distance between the central maximum and the $n$ th bright fringe is given by the formula:

$y_n = \frac{n\lambda L}{d}$

where $d$ is the distance between the two slits.

We can rearrange this formula to solve for $d$ :

$d = \frac{n\lambda L}{y_n}$

Substituting the values given in the question, we get:

$d = \frac{5 \times 600 \times 10^{-9} \times 1 \times 100}{5}$

$d = 6 \times 10^{-5} \text{ m}$

$d = 60 \ \mu\text{m}$

Therefore, the separation between the slits is $60 \ \mu\text{m}$ .