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 :

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, $y = 5 \text{ cm} = 0.05 \text{ m}$. <br/><br/> The distance between the central maximum and the $n$th bright fringe is given by the formula: <br/><br/> $y_n = \frac{n\lambda L}{d}$ <br/><br/> where $d$ is the distance between the two slits. <br/><br/> We can rearrange this formula to solve for $d$: <br/><br/> $d = \frac{n\lambda L}{y_n}$ <br/><br/> Substituting the values given in the question, we get: <br/><br/> $d = \frac{5 \times 600 \times 10^{-9} \times 1 \times 100}{5}$ <br/><br/> $d = 6 \times 10^{-5} \text{ m}$ <br/><br/> $d = 60 \ \mu\text{m}$ <br/><br/> Therefore, the separation between the slits is $60 \ \mu\text{m}$.