1) A thin film suspended in air is 0.410 µm thick and is illuminated with white light incident perpendicularly on its surface. The index of refraction of the film is 1.54. At what wavelength will visible light that is reflected from the two surfaces of the film undergo fully constructive interference?
2) Monochromatic green light, of wavelength 550 nm, illuminates two parallel narrow slits 7.70 μm apart. Calculate the angular deviation (θ) of the third-order (m = 3) bright fringe (a) in radians and (b) in degrees.
3) . How much faster, in meters per second, does light travel in a crystal with refraction index 1.77 than in another with refraction index 2.42?
1) using, t = n*lamda/(2*mue) (n = 1,2,3......)
==> lamda = 2*mue*t/n
= 2*1.54*0.41*10^-6/n
= 1.26*10^-6/n m
if n = 2
lamda = 1.26*10^-6/2 m
= 630 nm <<<<<<<<----------------------Answer
of n = 3
lamda = 1.26*10^-6/3
= 420 nm <<<<<<<<----------------------Answer
2) for third order, d*sin(theta) = 3*lamda
sin(theta) = 3*lamda/d
sin(theta) = 3*550*10^-9/(7.7*10^-6)
theta = 12.4 degree (or) 0.216 radians <<<<<<<<----------------------Answer
3) we know, v = c/n
v1 = c/n1 = 3*10^8/1.77 = 1.69492*10^8 m/s
v2 = c/n2 = 3*10^8/2.42 = 1.2397*10^8 m/s
so, v1 - v2 = 1.69492*10^8 - 1.2397*10^8
= 4.55*10^7 m/s <<<<<<<<----------------------Answer
1) A thin film suspended in air is 0.410 µm thick and is illuminated with white...
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