Q 12-10-126JEE MainJEE Main 2020 (7 Jan, Shift 2)Easy
In a Young's double slit experiment, the separation between the slits is $0.15$ mm. In the experiment, a source of light of wavelength $589$ nm is used and the interference pattern is observed on a screen kept $1.5$ m away. The separation between the successive bright fringes on the screen is:
Answer: (C) $5.9$ mm
$$\beta = \frac{\lambda D}{d} = \frac{589\times10^{-9}\times1.5}{0.15\times10^{-3}} = 5.89\times10^{-3}\ \text{m} \approx 5.9\ \text{mm}$$
Solution by Sreeraj P, M.Sc Physics