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Red light of wavelength 675 nm is incident on a slit of width 4.56 × 10−6...

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Physics

Red light of wavelength 675 nm is incident on a slit of width4.56 × 10−6 m. An observing screen is placed 1.50 m from theslit.

9a) Find the distance between the third order dark fringe andthe central bright fringe (in meters).

9b) If you replaced the single slit with two slits centered onthe former position of the single slit, what would the separationbetween the two slits need to be in order for the second orderbright fringe of the double-slit interference pattern to fall onthe third order dark fringe of the single slit diffractionpattern?

9c) If the average adult pupil diameter is 3.5 mm, what distancefrom the observing screen is a person no longer able to resolve(distinguish) the second order bright fringe and the central brightfringe in the double slit interference pattern?

Answer & Explanation Solved by verified expert
4.0 Ratings (431 Votes)

9a) lamda = 675 nm
a = 4.56*10^-6 m
L = 1.50 m
the distance between the third order dark fringe and the central bright fringe,
y3 = 3*lamda*L/a

= 3*675*10^-9*1.5/(4.56*10^-6)

= 0.666 m <<<<<<<<<<<-----------------------Answer


9b) let d is the slit separation.

for 3rd rark fringe, y3 = (2.5)*lamda*L/d

d = 2.5*lamda*L/y3


d = 2.5*675*10^-9*1.50/0.666

= 3.80*10^-6 m <<<<<<<<<<<-----------------------Answer


9c) minimum angular separtion to resolve, theta = 1.22*lamda/d

= 1.22*675*10^-9/(3.5*10^-3)

= 2.35*10^-4 radians

distance between central bright fringe and 2nd bright fringe, y2 = 2*lamda*L/d

= 2*675*10^-9*1.5/(3.8*10^-6)

= 0.5329 m
let r is the minimum distance between the person and the screen.

use, y2 = r*theta

r = y2/theta

= 0.5329/(2.35*10^-4)

= 2267 m <<<<<<<<<<<-----------------------Answer
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