The box of a pin hole camera, of length L, has a hole of radius a. It is assumed that when the hole

Q: Why do we just add the geometrical and diffraction spreads linearly?

In reality, the exact intensity distribution is a complex convolution of the geometrical shadow and the Airy disk diffraction pattern. However, for a simplified analytical model (which is standard in JEE), adding the characteristic widths of both effects gives a very good approximation of the total spot size.

Q: Why is the diffraction spread $L\lambda/a$?

The first minimum of a single slit (or circular aperture, ignoring the 1.22 factor for simplicity as the problem implies) occurs at an angle $\theta$ where $\sin\theta = \lambda/a$. For small angles, $\sin\theta \approx \theta$. The linear spread on a screen at distance $L$ is then $L\tan\theta \approx L\theta = L\lambda/a$.

Q: Why did we use the radius instead of the diameter for the spread?

The problem defines the 'spread' as the sum of the geometrical spread and diffraction spread. If we take the geometrical spread as the radius $a$, then the diffraction spread is the angular distance from the center to the first minimum, which is also a 'radius' of the diffraction pattern. The math works out the same whether you use radius or diameter, as long as you are consistent.

Q: What happens if we use a lens instead of a pinhole?

A lens focuses the geometrical rays to a single point, effectively making the geometrical spread zero at the focal plane. Then, the only spread left is the diffraction spread, which is why telescopes with lenses or mirrors are limited only by diffraction (the Rayleigh criterion).

Q: Why is $\sin\theta \approx \theta$ valid here?

The wavelength of visible light $\lambda$ is on the order of $10^{-7}$ meters, while a typical pinhole radius $a$ is on the order of $10^{-3}$ meters. Thus, $\lambda/a$ is around $10^{-4}$, which is a very small number. For such small values, the sine of the angle is virtually identical to the angle itself in radians.

Question

The box of a pin hole camera, of length $L$, has a hole of radius $a$. It is assumed that when the hole is illuminated by a parallel beam of light of wavelength $\lambda$ the spread of the spot (obtained on the opposite wall of the camera) is the sum of its geometrical spread and the spread due to diffraction. The spot would then have its minimum size (say $b_{	ext{min}}$) when

Options:

(A) $a = \sqrt{\frac{\lambda L}{2}}$ and $b_{	ext{min}} = \left(\frac{2\lambda^2}{L}ight)$

(B) $a = \sqrt{\lambda L}$ and $b_{	ext{min}} = \left(\frac{2\lambda^2}{L}ight)$

(C) $a = \sqrt{\lambda L}$ and $b_{	ext{min}} = \sqrt{4\lambda L}$

(D) $a = \frac{\lambda^2}{L}$ and $b_{	ext{min}} = \sqrt{4\lambda L}$

Sigma Percentile · Accepted Answer

### The Pinhole Camera Dilemma

Imagine you are building a pinhole camera. You have a box of length $L$ and you poke a small hole of radius $a$ in the front. A parallel beam of light, perhaps from a distant star, enters this hole. If light were purely a stream of particles traveling in perfectly straight lines, the spot on the back wall of the camera would be a crisp, perfect circle of radius $a$. This is what we call the **geometrical spread**.

But the universe is more interesting than that. Light is a wave! When a wave passes through a tiny aperture, it doesn't just go straight through; it bends and spreads out. This phenomenon is known as **diffraction**.

### The Wave Nature of Light

Because of diffraction, the light beam spreads out into a cone. The angular spread $	heta$ from the center to the first minimum of the diffraction pattern is given by the classic single-slit equation:

$$\sin	heta = \frac{\lambda}{a}$$

Since the hole radius $a$ is very small, the angle $	heta$ is also very small. In physics, we love the small-angle approximation, which tells us that $\sin	heta \approx 	heta$. So, the angular spread is simply $	heta = \frac{\lambda}{a}$.

Over the length $L$ of the camera box, this angular spread translates into an additional linear spread on the screen. Using basic trigonometry (arc length = radius $	imes$ angle), this diffraction spread is $L	heta = \frac{L\lambda}{a}$.

### The Master Equation

The problem states that the total spread of the spot, let's call it $b$, is the sum of the geometrical spread and the diffraction spread.

$$b = a + \frac{L\lambda}{a}$$

This equation is beautiful because it highlights a fundamental trade-off. If you make the hole radius $a$ very large, the geometrical spread $a$ dominates, and the image is blurry. If you make the hole radius $a$ very small, the diffraction spread $\frac{L\lambda}{a}$ blows up, and the image is again blurry! There must be a "Goldilocks" zone—an optimal hole size that minimizes the total spread $b$.

### Calculus to the Rescue

To find this minimum spot size, we turn to calculus. We need to find the value of $a$ that minimizes the function $b(a)$. We do this by taking the derivative of $b$ with respect to $a$ and setting it to zero.

$$\frac{db}{da} = \frac{d}{da}\left(a + \frac{L\lambda}{a}ight) = 0$$

The derivative of $a$ is $1$, and the derivative of $\frac{L\lambda}{a}$ is $-\frac{L\lambda}{a^2}$.

$$1 - \frac{L\lambda}{a^2} = 0$$

Solving for $a$, we get:

$$\frac{L\lambda}{a^2} = 1 \implies a^2 = L\lambda \implies a = \sqrt{\lambda L}$$

This is the optimal hole radius!

### Final Calculation

Now, we need to find the actual minimum size of the spot, $b_{	ext{min}}$. We substitute our optimal $a$ back into the master equation for $b$.

$$b_{	ext{min}} = \sqrt{\lambda L} + \frac{L\lambda}{\sqrt{\lambda L}}$$

Notice that $\frac{L\lambda}{\sqrt{\lambda L}}$ simplifies to just $\sqrt{\lambda L}$.

$$b_{	ext{min}} = \sqrt{\lambda L} + \sqrt{\lambda L} = 2\sqrt{\lambda L}$$

To match the options provided in the question, we can bring the $2$ inside the square root. Since $2 = \sqrt{4}$, we get:

$$b_{	ext{min}} = \sqrt{4\lambda L}$$

This perfectly matches option (c). The physics of the pinhole camera elegantly demonstrates how the wave nature of light imposes fundamental limits on optical resolution.

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Visualized Solution

$Geometrical Spread$

$Diffraction Spread$

$Linear Diffraction Spread$

$Total Spread b$

$Minimizing b$

$Differentiating b$

$Optimal Hole Radius a$

$Calculating b_{min}$

$Final Answer$

$The Trade-off$

The Sigma Insight: Diffraction due to Single Slit

The Pinhole Camera Dilemma

The Wave Nature of Light

The Master Equation

Calculus to the Rescue

Final Calculation

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Visualized Solution

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

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

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

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Optimal Hole Radius a

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

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

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