Q 12-10-150JEE MainJEE Main 2017 (2 Apr)Medium
In a Young's double slit experiment, slits are separated by $0.5$ mm, and the screen is placed $150$ cm away. A beam of light consisting of two wavelengths, $650$ nm and $520$ nm, is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is:
Answer: (C) $7.8$ mm
Bright fringes coincide where $n_1\lambda_1 = n_2\lambda_2$:
$$\frac{n_1}{n_2} = \frac{520}{650} = \frac45$$
The first coincidence is the $4$th bright fringe of $650$ nm (the $5$th of $520$ nm):
$$y = \frac{n_1\lambda_1D}{d} = \frac{4\times650\times10^{-9}\times1.5}{0.5\times10^{-3}} = 7.8\times10^{-3}\ \text{m} = 7.8\ \text{mm}$$
Solution by Sreeraj P, M.Sc Physics