JEE PYQ: Wave Optics Question 9
Question 9 - 2021 (26 Feb Shift 1)
In a Young’s double slit experiment two slits are separated by 2 mm and the screen is placed one meter away. When a light of wavelength 500 nm is used, the fringe separation will be :
(1) 0.75 mm
(2) 0.50 mm
(3) 1 mm
(4) 0.25 mm
Show Answer
Answer: (4)
Solution
Fringe width $(\beta) = \frac{\lambda D}{d}$
$d = 2 \times 10^{-3}$ m
$\lambda = 500 \times 10^{-9}$ m
$D = 1$ m
$\beta = \frac{500 \times 10^{-9} \times 1}{2 \times 10^{-3}}$
$\beta = \frac{5}{2} \times 10^{-4}$
$\beta = 2.5 \times 10^{-4}$
$\beta = 0.25$ mm
BETA
