Q 12-10-127JEE MainJEE Main 2020 (2 Sep, Shift 1)Easy
Interference fringes are observed on a screen by illuminating two thin slits $1\ \text{mm}$ apart with a light source $(\lambda = 632.8\ \text{nm})$. The distance between the screen and the slits is $100\ \text{cm}$. If a bright fringe is observed on a screen at distance of $1.27\ \text{mm}$ from the central bright fringe, then the path difference between the waves, which are reaching this point from the slits is close to:
Answer: (A) $1.27\ \mu\text{m}$
Path difference at height $y$ on the screen:
$$\Delta = \frac{yd}{D} = \frac{1.27\times10^{-3}\times1\times10^{-3}}{1} = 1.27\times10^{-6}\ \text{m} = 1.27\ \mu\text{m}$$
(This is about $2\lambda$, consistent with a bright fringe.)
Solution by Sreeraj P, M.Sc Physics