Question

Light of wavelength 630 nm falls on two slits and produces an interference pattern in which the third-order bright red fringe is 35 mm from the central fringe on a screen 2.8 m away. Part A What is the separation of the two slits?

Answer 1

Answer:

The separation of the two slits is = 1.5 ×$10^{-4}$ m

Explanation:

Wave length $\lambda$ = 630 nm = 0.63 mm

x = 35 mm

l = 2.8 m

Order of fringes (m) = 3

We know that the separation of the two slits is given by

$d = \frac{m \lambda \ l}{x}$

Put all the values in above formula we get

$d = \frac{(0.63)(3)(2.8)}{(35)}$

d = 1.5 ×$10^{-4}$ m

Therefore the separation of the two slits is = 1.5 ×$10^{-4}$ m