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. $



NCERT Chapter Video Solution

Dual Pane