A 0.60 mW laser produces a beam of light with a diameter of 1.3 mm . Part A What

Question

A 0.60 mW laser produces a beam of light with a diameter of 1.3 mm .

Part A

What is the average intensity of this beam?

Express your answer using two significant figures.

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Part B

At what distance does a 160 W lightbulb have the same average intensity as that found for the laser beam in part (a)? (Assume that 5.0% of the bulb's power is converted to light.)

Express your answer using two significant figures.

A 0.60 mW laser produces a beam of light with a diameter of 1.3 mm .

Part A

What is the average intensity of this beam?

Express your answer using two significant figures.

SubmitHintsMy AnswersGive UpReview Part

Part B

At what distance does a 160 W lightbulb have the same average intensity as that found for the laser beam in part (a)? (Assume that 5.0% of the bulb's power is converted to light.)

Express your answer using two significant figures.

Iav = kW/m2 r = cm  

Explanation / Answer

Part (A)

If a point source is radiating energy in three dimensions and there is no energy lost to the medium, then the intensity decreases in proportion to distance from the object squared. This is due to physics and geometry. Physically, conservation of energy applies. The consequence of this is that the net power coming from the source must be constant, thus:

P = integral (I.dA)

where P is the net power radiated, I is the intensity as a function of position, and dA is a differential element of a closed surface that contains the source. That P is a constant. If we integrate over a surface of uniform intensity I, for instance over a sphere centered around a point source radiating equally in all directions, the equation becomes:

P = I*A

= I*4*phi*R2

where I is the intensity at the surface of the sphere, and r is the radius of the sphere. (Asurf.= 4*phi*r2 is the expression for the surface area of a sphere). Solving for I, we get:

I= P/A

there fore I in first case = (0.60*10-3 )W /(3.14)*(0.65*10-3)2

I = 452.27 W/m2

Part (B)

We can treat the light bulb as a point source of light which radiates equally in all direction. Since it puts out 160 W of power, and light intensity is measured in W/m^2, we need to find the area through which the light is passing.

Since it is going in all directions, the area is that of a sphere with a radius equal to the distance from the light bulb. The formula for the area of a sphere of radius " R " meters is:

A = 4*phi*R^2

this implies intensity = 160/4*3.14*R^2

therefore, 452.27 = 160/4*3.14*R^2

this implies R^2 = 12.7388m2

hence R = 3.57 m

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