JEE PYQ: Wave Optics Question 20
Question 20 - 2020 (07 Jan Shift 2)
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:
(1) 6.9 mm
(2) 3.9 mm
(3) 5.9 mm
(4) 4.9 mm
Show Answer
Answer: (3)
Solution
Given, distance between screen and slits, $D = 1.5$ m
Separation between slits, $d = 0.15$ mm
Wavelength of source of light, $\lambda = 589$ nm
Fringe-width, $w = \frac{D}{d}\lambda = \frac{1.5}{0.15 \times 10^{-3}} \times 589 \times 10^{-9}$ m
$= 589 \times 10^{-2}$ mm $= 5.89$ mm $\approx 5.9$ mm
BETA
