Q 12-10-107JEE MainJEE Main 2021 (31 Aug, Shift 2)Easy
In a Young's double slit experiment, the slits are separated by $0.3$ mm and the screen is $1.5$ m away from the plane of slits. Distance between fourth bright fringes on both sides of central bright fringe is $2.4$ cm. The frequency of light used is $x\times10^{14}$ Hz.
Numerical value type. Enter your answer.
Answer: 5
Distance between the fourth bright fringes on both sides $= 8\beta = 2.4$ cm, so $\beta = 0.3$ cm.
$$\lambda = \frac{\beta d}{D} = \frac{3\times10^{-3}\times0.3\times10^{-3}}{1.5} = 6\times10^{-7}\ \text{m}$$
$f = \dfrac{c}{\lambda} = \dfrac{3\times10^8}{6\times10^{-7}} = 5\times10^{14}$ Hz, so $x = 5$.
Solution by Sreeraj P, M.Sc Physics