Question
Hey, I just need some quick help with these (mildly detailed work much appreciated).
1. In Young's double-slit experiment, a laster light of wavelength 632.8nm passes through two parallel slits and falls on a screen 10.5m away. Interference maxima on the screen are spaced 4cm apart. What is the distance between the slits?
2. A thin film of soap solution, n=1.33m with air on either side is illuminated normally with white light. Interference minima occur in the reflected light only at wavelengths of 400,480, and 600nm. What is the thickness of the film?
Thanks!!!!!
Explanation / Answer
from Young's double-slit experiment, y = L/d the distance between the slits is
d = L/y ............... (1) where, wavelength =
632.8*10-9 m distance between screen and slits L
= 10.5 m Interference maxima on the screen
y = 4*10-2 m substitute the given data in eq (1), we get
the distance between the slits is d = 166.11*10-6 m ................................................................................................. ................................................................................................. refractive index of soap solution
nsoap = 1.33 from constructive interference in thin films,
2nt = (m+1/2) ............. (1) for minimum :
m = 0 2nt = (1/2) thickness t = /4n ............. (2) minimumthickness of the oil film is
tmin = /4nsoap ........... (3) a) if wavelength = 400 nm , tmin = /4nsoap = (400 nm)/(4)(1.33)
= 75.18 nm b) if wavelength
= 480 nm ,
tmin = /4nsoap = (480 nm)/(4)(1.33) =
90.22 nm c) if wavelength
= 600 nm , tmin = /4nsoap = (600 nm)/(4)(1.33)
= 112.78 nm refractive index of soap solution
nsoap = 1.33 from constructive interference in thin films,
2nt = (m+1/2) ............. (1) for minimum :
m = 0 2nt = (1/2) thickness t = /4n ............. (2) minimumthickness of the oil film is
tmin = /4nsoap ........... (3) a) if wavelength = 400 nm , tmin = /4nsoap = (400 nm)/(4)(1.33)
= 75.18 nm b) if wavelength
= 480 nm ,
tmin = /4nsoap = (480 nm)/(4)(1.33) =
90.22 nm c) if wavelength
= 600 nm , tmin = /4nsoap = (600 nm)/(4)(1.33)
= 112.78 nm refractive index of soap solution
nsoap = 1.33 from constructive interference in thin films,
2nt = (m+1/2) ............. (1) for minimum :
m = 0 2nt = (1/2) thickness t = /4n ............. (2) minimumthickness of the oil film is
tmin = /4nsoap ........... (3) a) if wavelength = 400 nm , tmin = /4nsoap = (400 nm)/(4)(1.33)
= 75.18 nm b) if wavelength
= 480 nm ,
tmin = /4nsoap = (480 nm)/(4)(1.33) =
90.22 nm c) if wavelength
= 600 nm , tmin = /4nsoap = (600 nm)/(4)(1.33)
= 112.78 nm = (400 nm)/(4)(1.33)
= 75.18 nm b) if wavelength
= 480 nm ,
tmin = /4nsoap = (480 nm)/(4)(1.33) =
90.22 nm c) if wavelength
= 600 nm , tmin = /4nsoap = (600 nm)/(4)(1.33)
= 112.78 nm tmin = /4nsoap = (480 nm)/(4)(1.33) =
90.22 nm c) if wavelength
= 600 nm , tmin = /4nsoap = (600 nm)/(4)(1.33)
= 112.78 nm c) if wavelength
= 600 nm , tmin = /4nsoap = (600 nm)/(4)(1.33)
= 112.78 nm if wavelength
= 600 nm , tmin = /4nsoap = (600 nm)/(4)(1.33)
= 112.78 nm