Sunlight of intensity 1.3 kWm^-2 is incident normally on a thin convex lens of focal length 20 cm. I

Q: Why does the intensity change if no energy is lost?

Intensity is power per unit area. While the total power (energy per second) remains constant, the area over which this power is distributed changes as the beam converges and diverges. A smaller area means a higher intensity.

Q: Why do we use similar triangles here?

The light rays travel in straight lines from the edge of the lens to the focal point, and then continue in straight lines. This forms two cones joined at the apex (the focal point). Taking a cross-section gives us similar triangles.

Q: What happens exactly at the focal point (20 cm)?

According to ray optics, the area becomes zero, which would make the intensity infinite. However, in reality, wave optics (diffraction) prevents the beam from focusing to an infinitely small point, so the intensity reaches a very high, but finite, maximum value.

Q: Does the refractive index of the lens matter in this calculation?

Not directly for this part of the calculation. The refractive index and the curvature of the lens determine the focal length (20 cm). Since the focal length is already given, we don't need the refractive index.

Q: Why is the distance from the focus to the screen 2 cm?

The lens is at 0 cm. The focus is at 20 cm. The screen is at 22 cm. The distance from the focus to the screen is simply $22 \text{ cm} - 20 \text{ cm} = 2 \text{ cm}$.

Question

Sunlight of intensity $1.3 	ext{ kWm}^{-2}$ is incident normally on a thin convex lens of focal length 20 cm. Ignore the energy loss of light due to the lens and assume that the lens aperture size is much smaller than its focal length. The average intensity of light, in $	ext{kW m}^{-2}$, at a distance 22 cm from the lens on the other side is ............ .

Sigma Percentile · Accepted Answer

### The Concentrated Power of Sunlight

Imagine you are holding a magnifying glass on a bright, sunny day. You angle it just right, and the sunlight converges into a tiny, blindingly bright dot that can easily burn a hole through a piece of paper.

This classic childhood experiment is a perfect demonstration of the physics we are about to explore. The lens does not magically create more light or energy; it simply takes the energy spread over a large area and forces it into a much smaller one.

In our problem, sunlight with an initial intensity of $1.3 	ext{ kW/m}^2$ falls on a convex lens. The lens has a focal length of $20 	ext{ cm}$. We are asked to find the intensity of this light on a screen placed $22 	ext{ cm}$ away from the lens.

Notice that the screen is placed **past** the focal point. This means the light rays will converge to a point at $20 	ext{ cm}$, cross each other, and then begin to diverge again.

### The Master Principle: Conservation of Energy

The core principle governing this entire setup is the **Conservation of Energy**. Since the problem explicitly tells us to ignore any energy loss due to the lens, the total power of the light entering the lens must be exactly equal to the total power hitting the screen.

But what is power in this context? Power ($P$) is simply the product of the light's intensity ($I$) and the cross-sectional area ($A$) it covers.

Mathematically, we can express this conservation as:
$$P_{	ext{in}} = P_{	ext{out}}$$

Substituting our variables, we get:
$$I_0 A_0 = I' A'$$

Here, $I_0$ and $A_0$ are the initial intensity and area at the lens, while $I'$ and $A'$ are the final intensity and area at the screen. Our goal is to find $I'$.

### Unlocking the Geometry: Similar Triangles

To find the new intensity, we need to know how much the area has changed. Since the area of a circular beam depends on its radius ($A = \pi r^2$), we first need to find the ratio of the beam's radius at the screen to its radius at the lens.

Let's visualize the path of the light. The rays from the edge of the lens travel in straight lines to the focal point, forming a large cone. After crossing the focal point, they continue in straight lines, forming a smaller, inverted cone.

If we take a 2D cross-section of this setup, we see two **similar triangles** meeting at their apex (the focal point).

The large triangle (from the lens to the focus) has a height equal to the lens radius, let's call it $R$. Its length is the focal length, which is $20 	ext{ cm}$.

The small triangle (from the focus to the screen) has a height equal to the beam's radius at the screen, let's call it $r$. Its length is the distance from the focus to the screen. Since the screen is at $22 	ext{ cm}$ and the focus is at $20 	ext{ cm}$, this length is $22 - 20 = 2 	ext{ cm}$.

Because these triangles are similar, the ratio of their heights is equal to the ratio of their lengths:
$$\frac{r}{R} = \frac{2 	ext{ cm}}{20 	ext{ cm}}$$

Simplifying this fraction, we get:
$$\frac{r}{R} = \frac{1}{10}$$

### The Final Calculation

Now that we have the ratio of the radii, we can easily find the ratio of the areas. The area of a circle is proportional to the square of its radius.

Therefore, the ratio of the new area to the original area is:
$$\frac{A'}{A_0} = \left(\frac{r}{R}ight)^2$$

Substituting our radius ratio:
$$\frac{A'}{A_0} = \left(\frac{1}{10}ight)^2 = \frac{1}{100}$$

This tells us that the light beam at the screen is spread over an area that is $1/100$th the size of the lens.

Let's return to our master equation for power conservation:
$$I_0 A_0 = I' A'$$

We can rearrange this to solve for the final intensity, $I'$:
$$I' = I_0 \left(\frac{A_0}{A'}ight)$$

Since $A'/A_0 = 1/100$, the inverse ratio $A_0/A'$ is simply $100$.

Substituting the given initial intensity ($I_0 = 1.3 	ext{ kW/m}^2$):
$$I' = 1.3 	imes 100$$

$$I' = 130 	ext{ kW/m}^2$$

The average intensity of the light at a distance of $22 	ext{ cm}$ is $130 	ext{ kW/m}^2$. This problem beautifully demonstrates how a simple geometric analysis, combined with a fundamental conservation law, can easily solve what might initially seem like a complex optics question.

Enter Numerical Value:

Visualized Solution

$Visualizing the Setup$

$P_{in} = P_{out}$

$Area Ratio$

$\frac{r}{R} = \frac{2}{20}$

$\frac{A ^{'}}{A _{0}} = \frac{1}{100}$

$I^{'} = I_{0} \times 100$

$I^{'} = 130 kW/m^{2}$

The Sigma Insight: Lens

The Concentrated Power of Sunlight

The Master Principle

Unlocking the Geometry

The Final Calculation

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Enter Numerical Value:

Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Pin​=Pout​

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Rr​=202​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A0​A′​=1001​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I′=I0​×100

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I′=130 kW/m2