COMEDK 2025 Evening Solved Paper

Concept:The positions of bright and dark fringes in Young's double slit experiment depend on the wavelength of light, and the wavelength changes when light enters a medium of refractive index $\mu$.Explanation:In air, the wavelength is $\lambda$.The position of the $m$th bright fringe is given by $y_m^{\text{bright}} = m \frac{\lambda L}{d}$, where $d$ is the slit separation and $L$ is the screen distance.For the 4th bright fringe in air, $m = 4$, so $y_4 = 4 \frac{\lambda L}{d}$.In a medium of refractive index $\mu$, the wavelength reduces to $\lambda' = \frac{\lambda}{\mu}$.The position of the $n$th dark fringe in the medium is given by $y_n^{\text{dark}} = \left(n + \frac{1}{2}\right) \frac{\lambda' L}{d} = \left(n + \frac{1}{2}\right) \frac{\lambda L}{\mu d}$.For the 7th dark fringe in the medium, $n = 7$, so $y_7 = \left(7 + \frac{1}{2}\right) \frac{\lambda L}{\mu d} = \frac{7.5 \lambda L}{\mu d}$.The problem states that the 7th dark fringe in the medium lies at the same position as the 4th bright fringe in air.Equating the two positions:$\frac{7.5 \lambda L}{\mu d} = \frac{4 \lambda L}{d}$.Cancelling common factors $\lambda$, $L$, and $d$, we get $\frac{7.5}{\mu} = 4$.Solving for $\mu$: $\mu = \frac{7.5}{4} = 1.875$.Answer:The value of $\mu$ is $1.875$, which corresponds to Option C.