Light of wavelength 610 nm fell on a screen with two slilts that were 1.4 x 10 -

Question

Light of wavelength 610 nm fell on a screen with two slilts that were 1.4 x 10-4 m apart. A photographic plate is placed at a distance L from the slits. Alternating bright and dark frines form on the photographic plate. Third bright fringe (m=3) appears 0.15 m from the central bright fringe (m=0) ont eh photographic plate.

a) How far is the photographic plate from the slits?

b) What is the distance between first and second dark fringses on the photographic plate?

c) If you had to repeat the experiment with botht he slits and the photographic plate immersed in a liquid with index of refraction n=1.33, explain how would you modify the equation dsin(theta)=m(lamda) for Young's double-slits-experiment?

Explanation / Answer

for maximum in two slit

d*sintheta = m*lambda

tan theta = y/L

theta is small so tan theta = sin theta

d*y/L = m*lambda

m = 3

d = 1.4 x 10^-4 m

y = 0.15 m

L = d*y/(m*lambda)

L = 11.475 m

part b )

for minima

d*y/L = (m+1/2)*lambda

for 1st m = 0

for 2nd m =1

y = 3/2 * L*lambda/d - 1/2 * L*lambda/d

y = 0.05 m

part c )

d*sintheta = m*lambda/n

in water lambda = lambda_air/n

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