JEE PYQ: Wave Optics Question 10
Question 10 - 2020 (02 Sep Shift 1)
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 a 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) 1.27 $\mu$m
(2) 2.87 nm
(3) 2 nm
(4) 2.05 $\mu$m
Show Answer
Answer: (1)
Solution
Path difference, $\Delta P = d \sin\theta = d\theta$
$d =$ distance between slits $= 1$ mm $= 10^{-3}$ mm
$D =$ distance between the slits and screen $= 100$ cm $= 1$ m
$y =$ distance between central bright fringe and observed fringe $= 1.27$ mm
$\therefore \Delta P = \frac{dy}{D} = \frac{10^{-3} \times 1.270}{1} \text{ mm} = 1.27$ $\mu$m
BETA
