Question

In a Young’s double slit experiment, the distance between the two slits is 0.5 mm and the distance between the screen and the slits is 1 m. When a light of wavelength 500 nm is incident on the slits, what would be distance between the two second bright fringes?

(a) 1 mm

(b) 2 mm

(c) 3 mm

(d) 4 mm

This question was posed to me in an interview for internship.

I would like to ask this question from Young’s Experiment topic in portion Interference of Engineering Physics II

Answer 1

The correct choice is (b) 2 mm

Explanation: As we know, β = λD/d. In this question, we need to find 2 X β.

Here, d = 0.5 mm and D = 1 m

Therefore, β = 500 nm / 0.5 mm

= 1 mm

Now, Separation between the second bright fringe on both sides of the central maxima = 2 X 1 mm

= 2 mm.