Light intensity of a flashlight?

In summary, the problem is asking for the intensity of light as it hits a wall a distance L from the light bulb, given the wattage of the bulb and the angle of the conical beam. The intensity is calculated using the formula I = W/(pi*(Ltan(theta))^2). The question then asks for the amount of energy absorbed by the wall in a time Δt, which is simply calculated as W*Δt. The entire wall is considered the area of the absorbed energy, so the area part in the intensity formula cancels out.
  • #1
superbnchic
7
0

Homework Statement


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?


Homework Equations


I = Power/Area


The Attempt at a Solution


I = W/[area of cone] = W/(pi*Lcos(theta))

I'm having a bit of trouble understanding this problem! Please help!
Thank you!
 
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  • #2
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
Thank you!
so basically it would be
Intensity = power/area of circle?
Therefore, I = W/(pi*(Ltan(theta))^2)??
 
  • #4
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
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
Assuming the wall absorbs all the light, it's simply Power * Time. Remember, a watt is a joule per second.
 
  • #7
Oh, I see.
Therefore, the answer would JUST be
W*Δt
 
  • #8
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
Yup!
 
  • #10
Thank you SO much!
 

What factors affect the light intensity of a flashlight?

The light intensity of a flashlight can be affected by several factors, including the type and power of the bulb, the battery voltage, and the reflector design.

How do I measure the light intensity of a flashlight?

The light intensity of a flashlight can be measured using a specialized tool called a light meter. This device measures the amount of light emitted from the flashlight in units of lumens.

How does the distance from the flashlight affect its light intensity?

The further away an object is from the flashlight, the less intense the light will appear. This is due to the inverse square law, which states that the intensity of light decreases as the square of the distance from the source.

What is the difference between lumens and lux?

Lumens and lux are both units of measurement for light intensity, but they measure different things. Lumens measure the total amount of light emitted from a source, while lux measures the amount of light that falls on a specific area.

Can the light intensity of a flashlight be increased?

Yes, the light intensity of a flashlight can be increased by using a higher power bulb, a more powerful battery, or a better reflector design. However, it is important to note that increasing the light intensity may also decrease the battery life of the flashlight.

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