Optics Question 1099
Question: In Young’s double slit experiment, the slits are 0.5 mm apart and interference pattern is observed on a screen placed at a distance of 1.0 m from the plane containing the slits. If wavelength of the incident light is 6000 Å, then the separation between the third bright fringe and the central maxima is
[AMU 1995]
Options:
A) 4.0 mm
B) 3.5 mm
C) 3.0 mm
D) 2.5 mm
Show Answer
Answer:
Correct Answer: B
Solution:
Separation $ n^{th} $ bright fringe and central maxima is $ x _{n}=\frac{n\lambda D}{d} $ .
So, $ x _{3}=\frac{3\times 6000\times {{10}^{-10}}\times 1}{0.5\times {{10}^{-3}}}=3.5mm. $
BETA
