Question

Interference fringes are observed on a screen by illuminating two thin slits 1 mm apart with a light source ($\lambda$ = 632.8 nm). The distance between the screen and the slits is 100 cm. If a bright fringe is observed on a screen at distance of 1.27 mm from the central bright fringe, then the path difference between the waves, which are reaching this point from the slits is close to :

(1) 2.87

(2) 2 nm

(3) 1.27 $\mu$m

(4) 2.05 $\mu$m

Comments

Very lengthy question

Answer 1

Ans. (3) 1.27 $\mu$m

Sol. $\Delta P = \mathrm{d}\sin\theta$ 

$= d\theta$

$={\mathrm{d}y\over D}={10^{-3}\times1.270mm\over1m}=1.27 \mu m$

Comments

Very short
Please explain more