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).
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.
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?"Wow !
So we have someone rooting for option 4)
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} ##.
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.
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.
Yes, the intensity of light (power/area) is 1/16th of the original one, but the photocurrent is not determined by the light intensity alone. If light with the same intensity illuminates a bigger area the current will be larger.
See my post 28. When taught without much care for detail, a textbook might say the photocurrent response is proportional to the intensity of the light, but with more care for detail, the photocurrent is proportional to the incident power.
Your figure shows the dependence of the current from the frequency of light.
