JEE MAIN - Physics (2023 - 25th January Morning Shift - No. 6)
In Young's double slits experiment, the position of 5$$\mathrm{^{th}}$$ bright fringe from the central maximum is 5 cm. The distance between slits and screen is 1 m and wavelength of used monochromatic light is 600 nm. The separation between the slits is :
60 $$\mu$$m
48 $$\mu$$m
36 $$\mu$$m
12 $$\mu$$m
Explanation
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 $60 \ \mu\text{m}$.
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 $60 \ \mu\text{m}$.
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