Transmission Axis 0,-0 Transmission Axis en 2 25 Laser beam o . A helium-neon la
ID: 2041164 • Letter: T
Question
Transmission Axis 0,-0 Transmission Axis en 2 25 Laser beam o . A helium-neon laser emits a beam of unpolarized light with a maximum electrical field Emax of 1250 N/C and frequency 4.65x104 Hz (a) What is the average intensity lo of the beam? This beam passes through three polaroid filters, as shown in the figure (b) What is the intensity of the beam at point C? (c) If the beam lo is cylindrical with a diameter of 2.0 mm, how many photons are emitted each second? d) When a beam of this light is directed toward a double slit a fifth-order maximum (bright fringe) is produced at an angle of 0.737. At what angle the 8th dark fringe will occur? e) What thickness, t, should a thin layer of magnesium fluoride film (n-1.38), coated on a flint-glass lens n-1.61) have, if the reflection of the above light is to be suppressed? Draw a magnified sketch of reflected rays 1 and 2 to show path and phase difference at the applicable point/s Tranmission Axis )90 Polarer Polerizing Filter Poleriing FitExplanation / Answer
1)Given,
Em = 1250 N/C ; f = 4.65 x 10^14 Hz
q)We know that
S = Em^2/2cu0
S = 1250^2/(2 x 3 x 10^8 x 4 x 3.14 x 10^-7) = 2073.38 W/m^2
Hence, S = 2073.38 W/m^2
b)I1 = S/2 = 2073.38/2 = 1036.69 W/m^2
I2 = 1036.69 x cos^2(25) = 851.53 W/m^2
I3 = 851.53 x cos^2(90 - 25) = 152.09 W/m^2
Hence, I3 = 152.09 W/m^2
c)I = P/A
P = I A
P = 2073.38 x 3.14 x (1 x 10^-3)^2 = 0.0065 W
E = P t
E = 0.0065 x 1 s = 0.0065
lambda = c/f = 3 x 10^8/(4.65 x 10^14) = 6.45 x 10^-7 m
energy of a single photon
E = hc/lambda = 6.626 x 10^-34 x 3 x 10^8/(6.45 x 10^-7) = 3.1 x 10^-19 J
n = 0.0065/3.1 x 10^-19 = 2.1 x 10^16 photons
Hence, n = 2.1 x 10^16 photons
d)We know that
d sin(theta) = m lambda
d = 5 x 6.45 x 10^-7/sin(0.707) = 2.51 x 10^-4 m
theta-8 = sin^-1(8 x 6.45 x 10^-7/(2.51 x 10^-4)) = 1.178 deg
Hence, theta-8 = 1.178 deg
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