A monochromatic point source S radiating wavelength 6000 A, with power 2 W, an aperture A of diamete

Q: Why does the aperture act as a new source?

According to Huygens' principle, every point on a wavefront acts as a secondary source of spherical wavelets. When the light passes through the aperture, it spreads out spherically from that new position, so we must calculate the flux based on the distance from the aperture to the detector.

Q: Why do we multiply by the area of the aperture, but divide by the area of the sphere?

The source emits light in all directions (a full sphere of $4\pi r^2$). To find the flux (photons per unit area), we divide by the sphere's area. To find the total photons passing through a specific window (the aperture), we multiply that flux by the window's area.

Q: How does the concave lens change the problem?

A concave lens diverges the light rays. To an observer (or detector) on the other side, the light appears to be coming from a virtual image closer to the lens. We must use the lens formula to find this new virtual source position and calculate the new distances.

Q: Why doesn't the stopping potential change when the lens is inserted?

Stopping potential depends only on the maximum kinetic energy of the photoelectrons, which in turn depends strictly on the *wavelength* (or frequency) of the incident light and the work function of the material. The lens changes the *intensity* (number of photons), but not their individual energy.

Q: What does the $80\%$ transmission of the lens mean?

It means that $20\%$ of the photons are absorbed or reflected by the lens material itself. Only $80\%$ of the photons that hit the lens successfully pass through to the other side, reducing the final photocurrent.

Question

A monochromatic point source $S$ radiating wavelength $6000 	ext{ \AA}$, with power $2 	ext{ W}$, an aperture $A$ of diameter $0.1 	ext{ m}$ and a large screen $SC$ are placed as shown in figure. A photoemissive detector $D$ of surface area $0.5 	ext{ cm}^2$ is placed at the centre of the screen. The efficiency of the detector for the photoelectron generation per incident photon is $0.9$.
(a) Calculate the photon flux at (per unit area per unit time) the centre of the screen and the photocurrent in the detector.
(b) If the concave lens $L$ of focal length $0.6 	ext{ m}$ is inserted in the aperture as shown, find the new values of photon flux and photocurrent. Assume a uniform average transmission of $80\%$ from the lens.
(c) If the work function of the photoemissive surface is $1 	ext{ eV}$, calculate the values of the stopping potential in the two cases (without and with the lens in the aperture).

Sigma Percentile · Accepted Answer

### The Setup: A Journey of Photons

Imagine a glowing point source emitting light in all directions. This light travels through a small opening, an aperture, and finally hits a detector on a screen far away. Our mission is to track these photons, step by step, and find out how many actually knock out electrons at the detector.

### Step 1: The Energy of a Single Photon

First, let's find the energy of a single photon. Using the formula $E = \frac{hc}{\lambda}$, we plug in the wavelength of $6000 	ext{ \AA}$.

This gives us an energy of about $2.06 	ext{ eV}$, which is $3.3 	imes 10^{-19} 	ext{ J}$.

### Step 2: The Aperture as a Gateway

Since the source has a power of $2 	ext{ W}$, it's emitting $2 	ext{ J}$ of energy every second. Dividing this total energy by the energy of one photon tells us that the source emits a massive $6.06 	imes 10^{18}$ photons per second!

Now, these photons spread out spherically. By the time they reach the aperture at $0.6 	ext{ m}$, their density, or flux, is $n_1 / 4\pi r^2$.

Multiplying this flux by the area of the aperture gives us the number of photons actually passing through it: $1.052 	imes 10^{16}$ per second.

### Step 3: Reaching the Detector

Here is a crucial concept: the aperture now acts as a new, secondary source of light spreading out again. The detector is $5.4 	ext{ m}$ away from this aperture.

So, the photon flux at the detector is $n_3$ divided by $4\pi (5.4)^2$. This gives us $2.87 	imes 10^{13}$ photons per second per square meter.

To find the photocurrent, we multiply the flux by the detector's area to get the total photons hitting it. But remember, the efficiency is $0.9$, meaning only $90\%$ of these photons eject an electron.

Finally, multiplying the number of ejected electrons by the elementary charge gives us a tiny photocurrent of $2.07 	imes 10^{-10} 	ext{ A}$.

### Step 4: The Concave Lens Twist

Now for part B. We insert a concave lens at the aperture. This lens will diverge the light, creating a virtual image of the source.

Using the lens formula with $u = -0.6 	ext{ m}$ and focal length $f = -0.6 	ext{ m}$, we find the image distance $v$ is $-0.3 	ext{ m}$. The light now appears to come from this virtual image $S'$.

Because the light appears to come from $S'$, the new effective distance to the detector is $0.3 + 5.4 = 5.7 	ext{ m}$. Also, the lens only transmits $80\%$ of the light.

Calculating the new flux with these updated values, and repeating our photocurrent calculation, we get a reduced current of $1.483 	imes 10^{-10} 	ext{ A}$.

### Step 5: The Stopping Potential

Finally, for part C, we need the stopping potential. According to Einstein's photoelectric equation, the maximum kinetic energy is the photon energy minus the work function.

That's $2.06 - 1.0$, giving $1.06 	ext{ eV}$. So, the stopping potential is $1.06 	ext{ V}$.

Since the lens doesn't change the wavelength of the light, this potential remains exactly the same in both cases!

Visualized Solution

$Analyzing the Setup$

$Energy of a Single Photon$

$Total Photons Emitted per Second$

$Photons Intercepted by the Aperture$

$Photon Flux at the Detector$

$Photocurrent in the Detector$

$Inserting the Concave Lens$

$New Flux and Photocurrent$

$Stopping Potential$

The Sigma Insight: Photoelectric Effect

The Setup

Step 1

Step 2

Step 3

Step 4

Step 5

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Energy of a Single Photon

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Total Photons Emitted per Second

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Photons Intercepted by the Aperture

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Photon Flux at the Detector

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Photocurrent in the Detector

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Inserting the Concave Lens

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } New Flux and Photocurrent

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Stopping Potential