Photoelectric current and a convex lens

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  • #1
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Homework Statement


photoelectric current.jpg


Homework Equations




The Attempt at a Solution



The photoelectric current is directly proportional to the intensity of the light falling on it .

It will not depend on the focal length of the lens .

When the lens of half the diameter is used , intensity is halved . This means the current should also become half , but this is not an option .

I believe I am making some conceptual mistake .

Please point out the mistake .
 

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Answers and Replies

  • #2
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Well, consider the amount of sunlight striking the lens... it depends on how large the lens is (the area of the lens).
 
  • #3
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Well, consider the amount of sunlight striking the lens... it depends on how large the lens is (the area of the lens).

You are right .

When diameter is halved , area becomes one fourth and intensity should also become one fourth . Is option 2) correct ?
 
  • #5
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@Gene Naden , answer given is option 4) and the explanation given is that the intensity of the image is directly proportional to the area and inversely proportional to the square of the focal length . This does give option 4) .

Now I don't understand how the intensity of the image depends on the square of the focal length of the lens .
 
  • #6
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Well, I don't see that either. The number of photons striking the photosensitive surface should only depend on the diameter of the lens.
 
  • #7
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OK .

@Drakkith , @Charles Link , @ehild do you all agree that intensity of the image doesn't depend on the focal length of the lens and that option 2) is correct ?
 
  • #8
George Jones
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Now I don't understand how the intensity of the image depends on the square of the focal length of the lens .

Consider the situation where the photomaterial is placed between the lens and the focal point, and where the material is completely immersed in sunlight from the lens.

Now, without changing positions, replace the lens with a a lens that has equal diameter, but twice the focal length. More of the focused sunlight will miss this lens.

I do, however, find it a little difficult to go from the question's wording to this situation.
 
  • #9
Charles Link
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OK .

@Drakkith , @Charles Link , @ehild do you all agree that intensity of the image doesn't depend on the focal length of the lens and that option 2) is correct ?
The photosensor is assumed to be a finite size, but in order to get all of the energy collected by the lens on the photosensor, it may be necessary to reposition the photosensor so that it is in the focal plane of the lens in both cases. The photosensor is assumed to be large enough that the entire focused image fits inside of the photosensor. Option (2) is correct. ## \\ ## Edit: If the photosensor is not repositioned, then its size could effect the results considerably. If it is quite small, it will only get all of the light from lenses if it is in the focal plane in both cases. (That really IMO should have been stated in the statement of the problem).
 
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  • #10
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Yes, I assumed that the photosensor is larger than the focused image.
 
  • #11
George Jones
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The photosensor is assumed to be large enough that the entire focused image fits inside of the photosensor.

Yes, I assumed that the photosensor is larger than the focused image.

Yes, this is the way I read the question. In my previous post, my intention was to illustrate an alternate situation in which the photocurrent is dependent on focal length.
 
  • #12
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Does everyone agree that intensity of the image is independent of the focal length of the lens ?
 
  • #13
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I think the intensity of the image may depend on the focal length. However, I still think that the current only depends on the diameter of the lens.
 
  • #14
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I think the intensity of the image may depend on the focal length. However, I still think that the current only depends on the diameter of the lens.

If intensity depends on the focal length then current also depends on the focal length . Isn't it ?
 
  • #15
Charles Link
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Does everyone agree that intensity of the image is independent of the focal length of the lens ?
The terminology is somewhat vague when you say "intensity of the image". The photosensor will normally respond to power collected with a photocurrent proportional to power collected in a linear manner. The actual watts/m^2 that result in the image formation doesn't matter. In the case of a focal length that is doubled, the focused image area will be 4x as big in the second case, because of the longer focal length. That does not affect the photocurrent. What does affect the photocurrent is the diameter of the lens, because it affects the power collected. A lens with one half the diameter will collect 1/4 as much power. ## \\ ## With the image area in the second case being 4x as large, and 1/4 as much power collected, the irradiance (watts/m^2) of the second image will be 1/16 that of the first image. (The watts/m^2 (irradiance) can be called the "intensity" of the image). If the photosensor was 1000 or more tiny pixels, (like that of the focal plane array in a camera), each pixel (that was inside the focused image) in the second case would generate 1/16 the photocurrent that it did in the first case. ## \\ ## Edit: Photosensors that produce photocurrent can be quite microscopic, such as pixels in the FPA (focal plane array) of a camera, or they can be quite large, such as having area A= 1 cm^2 is more. I'm assuming the one in this problem is of the larger variety.
 
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  • #17
Drakkith
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Does everyone agree that intensity of the image is independent of the focal length of the lens ?

No. Halving the lens diameter reduces the total amount of light by 4x. Doubling the focal length of the lens increases the size of the image by 2x, reducing the power of the light per unit area by 4x. So that's a 16x reduction in power per unit area. If 'intensity' means power per unit area then the intensity falls to 1/16th.

In the case of a focal length that is doubled, the focused image area will be 4x as big in the second case, because of the longer focal length. That does not affect the photocurrent.

That's the part that's a bit vague to me. I assume the question is trying to get across that the intensity of the light falls off to 1/16th for the reasons we mentioned, but that obviously depends on the size of the photodetector and other aspects of the experiment.
 
  • #18
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Wow !

Finally there is someone rooting for option 4) :smile:
 
  • #19
Charles Link
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Wow !

So we have someone rooting for option 4) :smile:
See the edited additions to my post 15. They could have been more descriptive in the statement of the problem. Did they mean "with a small pixel photosensor?"
 
  • #20
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If 'intensity' means power per unit area then the intensity falls to 1/16th.

Are there two different definitions of intensity OR is intensity used differently in different contexts ?

As far as I know intensity is power per unit area .

Doesn't current also become 1/16th of the former case ?
 
  • #21
Charles Link
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Are there two different definitions of intensity OR is intensity used differently in different contexts ?
Intensity normally does not refer to total collected power. It often is used loosely to refer to irradiance ## E ## (watts/m^2). Intensity in radiometry is actually intensity ##I ## watts/steradian, and typically refers to a point source radiating with an inverse square law irradiance, with ## E=\frac{I}{s^2} ##.
 
  • #22
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Intensity normally does not refer to total collected power. It often is used loosely to refer to irradiance ## E ## (watts/m^2). Intensity in radiometry is actually intensity ##I ## watts/steradian, and typically refers to a point source radiating with an inverse square law irradiance, with ## E=\frac{I}{s^2} ##.

Intensity is power/area . Due to decrease in area of lens , power becomes 1/4th . Due to increase in focal length , image becomes double in size or area of the image is 4 times the previous case . This means intensity becomes 1/16th of the previous case .

Please let me know precisely what is your objection to the above reasoning .
 
  • #23
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Doubling the focal length of the lens increases the size of the image by 2x

How do you infer this ?

Using the lens formula 1/v -1/u = 1/f and putting u=∞ , v = f .

Magnification in case of thin lens = v/u .

How did you get the result mathematically that image area is proportional to square of the focal length of the lens ?
 
  • #24
ehild
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The photocurrent I is not the same as the intensity of light. As far as I know the photocurrent is about proportional to the power of light falling on the active area of the photocell. If the area is big enough so the whole light going through the lens falls on it, then the focal length does not influence the current. If the photocell is not adjusted to the new focal length, or its area is smaller then the image, the current becomes smaller than 1/4 of the original current.
 
  • #25
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The photocurrent I is not the same as the intensity of light.

Yes . But the photocurrent is proportional to the intensity of light .

then the focal length does not influence the current

Doesn't the image becomes larger which means the power entering through the lens get distributed on larger area .

Intensity is power/ area .

Doesn't area on which light is incident affect the intensity and in turn affect the current ?
 
  • #26
ehild
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How do you infer this ?

Using the lens formula 1/v -1/u = 1/f and putting u=∞ , v = f .

Magnification in case of thin lens = v/u .

How did you get the result mathematically that image area is proportional to square of the focal length of the lens ?
The image size/object size is equal to the image distance/object distance. If the focal length is doubled, the image size is doubled, as the object distance and size are the same. The size means linear size, if it is doubled, the area becomes 4 times bigger.
 
  • #27
ehild
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Yes . But the photocurrent is proportional to the intensity of light .



Doesn't the image becomes larger which means the power entering through the lens get distributed on larger area .

Intensity is power/ area .

Doesn't area on which light is incident affect the intensity and in turn affect the current ?
The amount of photons entering into the material of the photocell (a semiconductor) generate free carriers which make the electric current. And current is the amount of charge going across the area in unit time.
 
  • #28
Charles Link
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Most/many photosensors have the property that you get nearly one electron per photon, (sometimes the quantum efficiency may be in the 60%-80% range), which means you get an amount of electrical charge proportional to the incident energy, because each photon has a specific energy. ## \\ ## An electrical charge per given energy is the same as a photocurrent that is proportional to the incident power, if you divide both by time. (Photosensor response is sometimes specified as ## R\, (Coulombs/Joule) ## or equivalently as ## R \, (Amperes/watt) ##).
 
  • #29
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The amount of photons entering into the material of the photocell (a semiconductor) generate free carriers which make the electric current

OK .

Do you agree that intensity of light falling on the metal ( placed at the focal plane ) becomes 1/16th in the latter case ?
 
  • #30
Charles Link
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If I may answer the previous question as well, yes, the intensity is 1/16.
 

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