Question

12. Two slits separated by a distance of 1mm are
illuminated with red light of wavelength
6.5 x 10-7 m. The interference fringes are observed
on a screen placed 1m from the slits. The distance
between the third dark fringe and the fifth bright
fringe on the same side of central maxima is
(1) 0.65 mm
(2) 1.62 mm
(3) 3.25 mm
(4) 4.88 mm

Answer 1

Answer:

(2) 1.62 mm

Explanation:

The formula for the double slit interference is

$y=\frac{m\lambda D}{a}$

where

y is the distance of the m-th order bright fringe from the central maximum, and it is also the distance between two consecutive bright fringes

$\lambda$ is the wavelength of the light

D is the distance of the screen from the slits

a is the width of the slits

In this problem,

$a = 1 mm = 0.001 m\\\lambda = 6.5\cdot 10^{-7}m\\D = 1 m$

Therefore, the distance between each bright fringe and the next one is

$y=\frac{(1)(6.5\cdot 10^{-7})(1)}{0.001}=6.5 \cdot 10^{-4} m = 0.65 mm$

Here we want to know the distance betweeh the 3rd dark fringe and the 5th bright fringe. In order from the closest to the farthest from the central maximum, we have:

- 3rd dark fringe

- 3rd bright fringe

- 4th dark fringe

- 4th bright fringe

- 5th dark fringe

- 5th bright fringe

The distance between a dark fringe and the next bright fringe is half the distance between two bright fringes: so, the the distance betweeh the 3rd dark fringe and the 5th bright fringe is 2.5 times the value we have calculated, therefore

$y' = 2.5 (0.65 mm)=1.62 mm$