Argon-ion laser light of free-space wavelength 488 nm is incident on the composi

Question

Argon-ion laser light of free-space wavelength 488 nm is incident on the composite medium of glass, water and air as shown in the figure above. The speed of light in free space is 3x108 m/s. Distance d along the water-air interference is measured from the vertical line in the middle.

a) What is the wavelength of this light in glass? What is its speed in water?

b) There is a target at d=6m. Find the value of for which the laser light will hit the target. Find the direction(s) if propagation of the laser light after it hits the water-air interface?

c) Assume that the incident beam is unpolarized. For what value will the reflected light from the glass-water interface be completely plane polarized? What will be the direction of polarization? An ideal polarizer is placed in the path of the reflected beam with its polarizing axis at an angle of 60° with the plane of incidence. If the reflected polarized beam has an intensity of 0.02 W/m2, what will be the intensity of the beam after it passes through the polarizer?

Explanation / Answer

a) wavelength in glass = 488 / 1.52 = 321 nm

wavelength in water = 488/1.33 = 367 nm

b) from the given picture, d = 6 c = 5 and tan theta = 6/5

so theta = 50.2 degrees

The angle of incidence on the water-air interface is 50.2 degrees, so the beam will internally reflect at at angle of 50.2 degrees above the horizontal.

c) this problem refers to "Brewsters angle", which is given by

tan theta = n2 / n1 = 1.33/1.52

so theta = 41.2 degrees

direction of polarization is parallel to the interface (i.e. in-out of the page)

intensity after polarizer = 0.02 * (cos60)^2 = 0.005 W/m^2

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