Laser light can permanently damage our eyes. Laser light\'s intensity, collimati

Question

Laser light can permanently damage our eyes. Laser light's intensity, collimation and directionality are the main reason that a 1 mW Helium-Neon (HeNe) laser can permanently damage your eyes while a 60 W light bulb is harmless to look at. This exercise lets you estimate and compare the intensities of 3 different light sources at the retina. Let us model the eye's cornea and lens as shown to the right with a typical distance of 20.0 mm from the lens to the retina. We assume a pupil opening of 2 mm in diameter. The lens focuses light onto a spot on the retina just like a converging lens, (magnifying glass) focuses sun's light to a spot to burn a piece of paper. Based on the following information, determine the intensity, (power per unit area, W/m2), at the retina for three different light sources of light bulb, the Sun, and a HeNe laser. 1) A 60 W incandescent light bulb at 1 m from the eye. Assume a spot size of about 1 mm in diameter is formed at the retina 2) Assume approximately 1000 W/m2 for the sun's intensity near our cornea, and a spot size of about 0.5 mm at the retina. 3) A 1.0 mW HeNe laser having a beam diameter of about 2 mm. This beam will be focused to a very small spot size of about 10 um in diameter. 4) Consider in the above Fig. 1, the above HeNe laser is incident from left on the 1 polarizer. We rotate the polarizer around z-axis and notice that the transmitted laser light gradually extinguishes when the polarizer's transmission axis is along x-axis (vertical). (a) What can we conclude about the polarization of the laser light source?Explain. (b) Next we set the transmission axis of the 1 polarizer at 35° with respect to x-axis. Calculate the intensity of the laser light before and after the polarizer. (c) Next, what should we set the angle of a 2d polarizer, (the analyzer), with respect tox-axis such that no light is passed through it? Explain, (d) At this setting, when no light is passed through the analyzer, we insert a 3 polarizer between the polarizer and the analyzer. Determine the angular range with respect tox-axis such that the 3e polarizer would cause light to be transmitted through the analyzer. Explain why this occurs and at what angle with respect tox-axis maximum light is transmitted through the analyzer? Retina Lens

Explanation / Answer

Part 1).

Power, P = 60 W

distance form the eye = 1 m

pupil opening, 2r = 2 mm

spot diameter on retina, d = 1 mm

Intensity, Ibulb = 4(P*pi*r2/4*pi*D2)/pi*d2

= (P*r2/D2)/pi*d2

Hence, intensity recieved at the retina due to light bulb is, Ibulb = 19098.59 W/m2

Part 2).

Sun's intensity, ISun = 1000 W/m2

sport size, d = 0.5 mm

hence

I' = I*(2r)2/d2 = 4I(r/d)2

Intensity recieved at the retina due to Sun is, I'Sun = 16000 W/m2

Part 3)

HeNe LASER Power, P = 1mW

beam diameter, d = 2 mm

spot diameter, d' = 10*10-6 m

hence,

intensity I = P/pi(d'/2)2

Intensity recieved at the retina due to HeNe LASER is = 1.27*107 W/m2

Part 4)

(a.)

As the light diminishes when the polariser is along x-axis, hence we conclude that the polarisation of incident beam is along y-axis, as transmitted light perpendicular to this polarisation (along x-axis) will be 0.

(b.)

phi = 90o - 35o = 55o

Let intensity of light before is 'Io' and intensity of light after is 'I'

I = Io*cos2 55o

I = 0.33 Io

(c.)

for the next polariser, let the angel with x-axis be 'phi'

then

phi = 35o +- 90o = 125o or - 55o

(d.)

when we introduce the third polariser, the range of angles for which light will be transmitted are such that the second polariser must not be perpendicular to the first or the third polariser. Also, the maximum light is transmitted when the second polariser is at half of the angle between the first and the third polariser.

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