# Homework Help: Light intensity of a flashlight?

1. Jul 22, 2009

### superbnchic

1. The problem statement, all variables and given/known data
A flashlight lights up a wall a distance L from the small bulb whose wattage is given by W. The conical beam emerges from the flashlight at an angle theta. What is the intensity of light as it hits the wall a distance L from the bulb?

2. Relevant equations
I = Power/Area

3. The attempt at a solution
I = W/[area of cone] = W/(pi*Lcos(theta))

Thank you!

2. Jul 22, 2009

### Pengwuino

The problem is basically asking hte amount of energy incident on the wall at the end of a light cone. The cone's base is the area that the light is incident upon. Now if you have a set distance L for the "height" of the cone, you can determine the radius of the base of that cone using the angle given. The base is a simple circle. Then you can use your intensity = power/area formula. By the way, your equation for the area of the cone is incorrect. Look at the dimensions, it also might help to draw a diagram for it.

3. Jul 22, 2009

### superbnchic

Thank you!
so basically it would be
Intensity = power/area of circle?
Therefore, I = W/(pi*(Ltan(theta))^2)??

4. Jul 22, 2009

### Pengwuino

Yup! Notice how as theta -> 0, the intensity blows up as expected if you make that cone smaller and smaller and smaller. Also as theta -> pi/2, the cone expands to being infinitely big so the intensity, that is power per area, drops to 0.

5. Jul 22, 2009

### superbnchic

Also,
there was another part to this question I had trouble understanding

Question
How much energy is absorbed by the wall in a time Δt?

attempt
Energy absorbed = Power * Δt
= [W/(pi*(Ltan(theta))^2)] * Δt

6. Jul 22, 2009

### Pengwuino

Assuming the wall absorbs all the light, it's simply Power * Time. Remember, a watt is a joule per second.

7. Jul 22, 2009

### superbnchic

Oh, I see.
Therefore, the answer would JUST be
W*Δt

8. Jul 22, 2009

### superbnchic

OH!
I get it.
What I was trying to do was
Energy absorbed = Intensity * area of the absorbed energy * Δt
But in this case, the entire wall is the area of the absorbed energy therefore the area part in the Intensity would cancel out thus simply giving me W*Δt
Is that correct?

9. Jul 22, 2009

### Pengwuino

Yup!

10. Jul 22, 2009

### superbnchic

Thank you SO much!!