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## Consider a source of light with wavelength lambda = 530 nm that shines on two identical narrow slits. The slits are separated by a distance

Question

Consider a source of light with wavelength lambda = 530 nm that shines on two identical narrow slits. The slits are separated by a distance a = 30 mu m. An interference pattern is observed on a screen located a distance L away from the slits. On the screen, the location of the second dark spot to the left of the central bright spot is found to be y = 1.2 cm from the central bright spot. Let this particular position on the screen be referred to as P_1. Light from both slits travels to the point P_1. How much further does the light from one slit travel compared to the light from the other slit? distance = 795 nm The path difference in part A is equal to a middot sin theta where the angle theta is the separation between the central bright spot and the second dark spot. What is this angle theta? theta = ___ degree Using the angle theta and the location y of the second dark spot on the screen, determine the distance L between the slits and the screen L = ____ m

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Physics
3 years
2021-07-26T17:18:39+00:00
2021-07-26T17:18:39+00:00 1 Answers
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## Answers ( )

Answer:Check the explanation

Explanation:A)

For destructive interference, path difference = n\lambda/2, n = 1, 3, 5 …

Where n = 1 corresponds to first dark spot, n = 3 corresponds the second dark spot and so on

So here, n = 3

distance = 3\lambda/2

= 1.5 x 530 nm

= 795 nm.

B)

Path difference, a sin\theta = 795 nm = 0.795 \mum

sin\theta = [0.795 \mum] / [30 \mum] = 0.0265

\theta = sin-1(0.0265)

= 1.52 degrees.

C)

tan\theta = y/L

L = y/tan\theta

= 1.2 cm / tan(1.52)

= 45.27 cm

= 0.45 m