Photoelectric current and a convex lens

[ehild]

OK .

Do you agree that intensity of light falling on the metal ( placed at the focal plane ) becomes 1/16th in the latter case ?

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.

 

[Jahnavi]

Please bear with me .

Is the textbook wrong when it says that the photocurrent is directly proportional to the intensity of light falling on the metal ?

 

[Charles Link]

Please bear with me .

Is the textbook wrong when it says that the photocurrent is directly proportional to the intensity of light falling on the metal ?

View attachment 226382

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.

 

[Jahnavi]

@Drakkith I might have misunderstood your post#17 .

Do you believe option 2) is correct ?

 

[ehild]

Please bear with me .

Is the textbook wrong when it says that the photocurrent is directly proportional to the intensity of light falling on the metal ?

View attachment 226382

Your figure shows the dependence of the current from the frequency of light.
And measuring how the current depends of the intensity of light, they might have made higher intensity by increasing power instead of decreasing the cross section of the light beam.

 

[ehild]

Anyway, the image size of the Sun is very small, much smaller than the size of a common photocell. So all the power going through the lens falls onto the photocell in both cases. If the photocell is in the focal point, the focal length should not count.

 

[Jahnavi]

Sorry .

Please see the fourth graph https://chem.libretexts.org/Textboo...f_Quantum_Mechanics/2.2:_Photoelectric_Effect

 

[ehild]

You can change the intensity of light either changing the power or by changing the cross-section of the beam. When they say intensity of light they usually think of power.

 

[kuruman]

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

Indeed. The intensity of light reaching the metal is reduced by a factor of 4 when the smaller lens is put in place because only the light that passes through the lens ejects electrons. This reduces the current by a factor of 4.
Stated differently: When put in place, the smaller lens gathers 1/4 of the Sun's photons that were gathered by the larger lens. This means that 1/4 of the electrons will be ejected from the surface per unit time.

 

[Jahnavi]

The intensity of light reaching the metal is reduced by a factor of 4

By a factor of 4 OR by a factor of 16 ?

Everyone in this thread agreed that light intensity reduces by a factor of 16 .

 

[ehild]

"Intensity of light" is used in different meanings. See https://en.wikipedia.org/wiki/Intensity_(physics)

 

[kuruman]

"Intensity of light" is used in different meanings. See https://en.wikipedia.org/wiki/Intensity_(physics)

Yes, that is why it is more productive to eliminate the middleman and think in terms of photons crossing the lens and striking the metal surface.

 

[Drakkith]

@Drakkith I might have misunderstood your post#17 .

Do you believe option 2) is correct ?

No, I think option 4 is correct.

 

[kuruman]

No, I think option 4 is correct.

Let $N_L$ = number of photons per unit time passing through the larger lens.
My argument is that
1. All the photons that pass through the larger lens hit the cathode.
2. The number of electrons ejected per unit time is $fN_L$ where $f$ is the cathode efficiency.
3. The current is proportional to the number of ejected electrons per unit time, $I_L=\alpha fN_L$.
4. The smaller lens gathers 1/4 of the photons gathered by the larger lens because its area is 1/4 of the larger lens area.
5. The number of photons per unit time passing through the smaller lens is $N_S=N_L/4$.
6. All the photons that pass through the smaller lens hit the cathode.
7. The number of electrons ejected per unit time is $fN_S=fN_L/4$.
8. The current is proportional to the number of ejected electrons per unit time, $fN_S$, $I_S=\alpha fN_S=\alpha fN_L/4$.
9. Therefore $I_S=I_L/4$.

Which of the statements 1-9 is fallacious and why? I don't see it.

 

[Drakkith]

1. All the photons that pass through the larger lens hit the cathode.

6. All the photons that pass through the smaller lens hit the cathode.

That's the key here I think. If all of the light that passes through the lens hits the cathode, regardless of the focal length, then the current will definitely be 1/4 and not 1/16.

 

[Jahnavi]

That's the key here I think. If all of the light that passes through the lens hits the cathode,

Why would that be not true ?

 

[Drakkith]

Why would that be not true ?

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.

 

[Jahnavi]

OK .

I have a very basic question . When the light intensity reduces to 1/16th , does the amplitude of the electromagnetic radiation becomes 1/4th ?

We have already discussed how intensity is power/area . But intensity is also proportional to square of the amplitude .

So , does the amplitude also decreases and becomes 1/4th ?

 

[Charles Link]

Yes, when the irradiance (watts/m^2) in the region of the focused image in the second case is 1/16 of the first case, the electric field amplitude in the second case will be 1/4 of the first case. $\\$ There is one extra detail here though I believe. For a monochromatic source, the irradiance is proportional to the square of the electromagnetic field amplitude. In the case of a broad spectrum source, I'm not sure the electric field amplitude is so readily defined. Even with just two wavelengths present, I don't know how the electric field would be determined. Each wavelength has an electric field amplitude associated with it. The irradiance for each wavelength is found by squaring the electric field amplitude for that wavelength.

 

[Jahnavi]

Thank you everyone for your valuable inputs .

It is really nice interacting and listening to so many experts

 

[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.