Photoelectric current and a convex lens

[kuruman]

It wouldn't be true if the image of the light source becomes larger than the cathode so that some of the light no longer falls on the photosensitive area and is lost.

I agree, however for the answer 1/16 to be correct, not some but exactly 1/4 of the light that makes it through the lens falls on the photosensitive area. The way I see it, the answer of 1/16 is obtained using the following reasoning
1. The larger lens is one focal length away from the photosensitive area.
2. With the larger lens in place, the Sun's image covers exactly the photosensitive area and there is no photosensitive material outside this area.
3. The smaller lens will gather only 1/4 of the photons that the larger lens gathered.
4. The smaller lens replaces the larger lens at the same distance from the photosensitive area.
5. The diameter of the image at the distance to the photosensitive area will double and the area of the image at that distance will quadruple.
6. Since the photosensitive area remains the same, but the area of the Sun's image is quadrupled, only 1/4 of the photons that make it through the smaller lens produce current, the overall factor being 1/4×1/4 = 1/16.

As @Drakkith pointed out, the choice of $(1/4)I$ is correct if one believes that a tacit assumption to the question is that all the photons that pass through either lens hit the photosensitive area.
The choice of $(1/16)I$ is correct if one believes that a tacit assumption to the question is statement 2 above.

Take your pick.

 

[Charles Link]

I agree, however for the answer 1/16 to be correct, not some but exactly 1/4 of the light that makes it through the lens falls on the photosensitive area. The way I see it, the answer of 1/16 is obtained using the following reasoning
1. The larger lens is one focal length away from the photosensitive area.
2. With the larger lens in place, the Sun's image covers exactly the photosensitive area and there is no photosensitive material outside this area.
3. The smaller lens will gather only 1/4 of the photons that the larger lens gathered.
4. The smaller lens replaces the larger lens at the same distance from the photosensitive area.
5. The diameter of the image at the distance to the photosensitive area will double and the area of the image at that distance will quadruple.
6. Since the photosensitive area remains the same, but the area of the Sun's image is quadrupled, only 1/4 of the photons that make it through the smaller lens produce current, the overall factor being 1/4×1/4 = 1/16.

As @Drakkith pointed out, the choice of $(1/4)I$ is correct if one believes that a tacit assumption to the question is that all the photons that pass through either lens hit the photosensitive area.
The choice of $(1/16)I$ is correct if one believes that a tacit assumption to the question is statement 2 above.

Take your pick.

The requirements for 1/16 to be correct are not so fussy. If the photocell is in the focal plane in both cases, and it is smaller than the first image, in both cases it will measure the irradiance level of the image which is a factor of 1/16 lower in the second case. (Power $P=E \, A_D$, where $E=$ irradiance level (watts/m^2). The photocurrent is proportional to the power $P$ which is also proportional to the irradiance level $E$. ) I'm assuming the sun has nearly uniform brightness across its image, which isn't precisely the case, but it is a reasonably good approximation. $\\$ I believe the problem statement needs to be more clear, because this second scenario, (of a photocell used to measure irradiance level as opposed to total collected power), can also be of interest in many cases. In a camera focal plane array of many, many pixels, this second scenario is how the pixels are employed. They are used to measure the brightness level of the image at the given location. With the additional assumption that the photosensor is very small, or even simply smaller than the first image, @Drakkith 's answer of 1/16 is, in fact, correct.

 

[ehild]

Assume a common lens with focal length of 1 m. What can be the image size of the Sun?

 

[Charles Link]

Assume a common lens with focal length of 1 m. What can be the image size of the Sun?

The size of an image in the focal plane $\Delta x=f \, \Delta \theta$. The sun subtends an angle, (when viewed from earth), just slightly less than $\Delta \theta=.01$ radians. With $f=1.0$ m , that means $\Delta x \approx 1.0$ cm. $\\$ Edit: A google shows the diameter of the sun, $d_{sun}=864,000$ miles, and $s_{earth-sun}=93,000,000$ miles. $\Delta \theta=\frac{d}{s}$.

 

[ehild]

As I remember, diameter of the cathode in the old-type vacuum photocells was about 1cm and the modern semiconductor photocells are of the same size or smaller. If we use a lens of f=1 m, the size of the image is about the same as that of the cathode. So the number of incident photons can ne proportional to the intensity of the image multiplied by the active area of the cell. So 1/16 wins, but using shorter focal lengths, or bigger photocathodes, both images can be smaller than the active area, than the result can be just the other one, 1/4.

 

[Tom.G]

Slightly off subject.

An octal base, type 930, photo detector by RCA. With a tube diameter of 1.3 inches, the photocathode area looks to be about 1 sq.in.
Photo from datasheet at: https://www.radiomuseum.org/tubes/tube_930.html

As a teenager, I put together a circuit using one of these. Don't recall why though; probably just because I could.

 

[Tirtha Biswas]

Hey, Jahnavi, I want to say something and don't believe me but give the proof to yourself. Let me consider the case of firing a paper using sunlight and convex lens. We know that intensity is inversely proportional to sq. Of distance. And the distance is nearly equal to the focal length since the distance from the sun is about infinity. So doubling focal length Doubles the distance; and intensity becomes one fourth.
Secondly the diameter is halved so the amplitude also becomes haft since it is directly proportional to diameter. And intensity being proportional to the square of amplitude, becomes one fourth.
One fourth into one fourth gives 1/16.
Hope that makes sense.