A biconvex lens of focal length f forms a circular image of radius r of sun in focal plane. Then whi

Q: Why is the image formed exactly at the focal plane?

Because the sun is effectively at infinity, the rays from any point on the sun are parallel to each other when they reach the lens. Parallel rays always converge at the focal plane of a lens.

Q: Why does covering half the lens not halve the image area?

Every single part of a lens forms the complete image. Covering half the lens just blocks half the light rays, making the image dimmer (halving the intensity), but it does not cut the image size in half.

Q: Why does intensity decrease when focal length is doubled?

Doubling $f$ doubles the radius $r$, making the area 4 times larger. The same amount of light energy is now spread over a 4 times larger area, so the intensity (energy per unit area) becomes 1/4th.

Q: Is the approximation $\tan \theta \approx \theta$ always valid?

It is highly accurate for very small angles measured in radians. The sun subtends an angle of about 0.5 degrees, which is roughly 0.009 radians, making the small angle approximation perfectly valid here.

Q: Does the refractive index of the lens affect the image size?

The refractive index affects the focal length $f$ of the lens. Once $f$ is determined, the image size depends only on $f$ and the angular size of the object $\theta$.

Question

A biconvex lens of focal length $f$ forms a circular image of radius $r$ of sun in focal plane. Then which option is correct ?

Options:

(A) $\pi r^2 \propto f$

(B) $\pi r^2 \propto f^2$

(C) If lower half part is convered by black sheet, then area of the image is equal to $\pi r^2 / 2$

(D) If $f$ is doubled, intensity will increase

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine the sun, located at a vast distance, sending its rays towards our biconvex lens. Because the sun has a finite size, these rays do not arrive perfectly parallel to the principal axis; instead, they arrive at a small angle $	heta$.

The ray passing through the optical center of the lens goes completely undeviated. It strikes the focal plane at a distance $r$ from the principal axis, which represents the radius of the sun's image.

### The Master Equation

Let's focus on the right-angled triangle formed by the optical center, the principal focus, and the edge of the image. The base of this triangle is the focal length $f$, and the height is the image radius $r$.

From this triangle, we can write the trigonometric relation:

$$ 	an 	heta = \frac{r}{f} $$

Since the sun is extremely far away, the angle $	heta$ is very small. For such small angles, $	an 	heta$ is approximately equal to $	heta$ itself. Therefore, we can write:

$$ r = f 	heta $$

### Final Calculation

This implies that the radius of the image is directly proportional to the focal length of the lens. Now, what about the area of this circular image?

The area of a circle is given by:

$$ A = \pi r^2 $$

Substituting our expression for $r$, we get:

$$ A = \pi (f 	heta)^2 $$

Since $	heta$ is a constant for the sun, the area is directly proportional to the square of the focal length:

$$ \pi r^2 \propto f^2 $$

So, the correct option is **(b)**.

### Exploring the Alternatives

Let's also quickly analyze the other options to understand why they are incorrect.

If we cover the lower half of the lens, the complete image is still formed because every part of the lens focuses light to create the entire image. So, the area remains the same, but the **intensity halves** since only half the light is collected.

And if we double the focal length, the area becomes four times larger. Since the total light energy collected by the lens remains the same, spreading it over a four times larger area means the **intensity will decrease** to one-fourth of its original value.

Animated Solution for Physics - Optics: A biconvex lens of focal length $f$ forms a circular image of radius $r$ of sun in focal plane. Then which option is correct ?

Select Answer:

Visualized Solution

$Visualizing the Setup$

$Image Formation$

$Geometric Relation$

$Small Angle Approximation$

$Radius Proportionality$

$Area Proportionality$

$Analyzing Other Options$

The Sigma Insight: Lens

Analyzing the Setup

The Master Equation

Final Calculation

Exploring the Alternatives

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Animated Solution for Physics - Optics: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A biconvex lens of focal length f forms a circular image of radius r of sun in focal plane. Then which option is correct ?