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 $$b_{min}$$) when:

We begin by analysing the two independent reasons for which the bright spot on the screen of a pin-hole camera acquires a finite size.

First reason (pure geometry). Rays that pass straight through the circular hole of radius $$a$$ reach the screen (placed at a distance $$L$$ behind the hole) without bending. Consequently the geometrical radius of the spot is exactly the same as the radius of the hole. Hence

$$\text{geometrical spread } = a.$$

Second reason (diffraction). Light emerging from the circular aperture behaves like light coming from a circular source; it diffracts and forms an Airy pattern. The linear radius of the central maximum on a distant screen is obtained from the simple Fraunhofer condition for the first minimum

$$\sin\theta \approx \theta = \frac{\lambda}{a},$$

so that for small angles the linear size on a screen a distance $$L$$ away is

$$\text{diffraction spread } = L\,\theta = L\left(\frac{\lambda}{a}\right)=\frac{\lambda L}{a}.$$

The two effects act independently along the same straight line, so the total radius $$b$$ of the spot is taken as the algebraic sum

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

Our task is to find the value of $$a$$ that minimises $$b$$. For this we differentiate $$b$$ with respect to $$a$$ and equate the derivative to zero.

First write the explicit expression again:

$$b(a)=a+\frac{\lambda L}{a}.$$

Now differentiate:

$$\frac{db}{da}=\frac{d}{da}\left(a\right)+\frac{d}{da}\left(\frac{\lambda L}{a}\right)=1-\frac{\lambda L}{a^{2}}.$$

The minimum occurs when

$$\frac{db}{da}=0\quad\Longrightarrow\quad 1-\frac{\lambda L}{a^{2}}=0.$$

Solving the above equation step by step, we obtain

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

$$a^{2}=\lambda L,$$

$$\boxed{a=\sqrt{\lambda L}}.$$

Substituting this optimal value of $$a$$ back into the formula for $$b$$ gives the minimum possible radius $$b_{\text{min}}$$:

$$\begin{aligned} b_{\text{min}} &=a+\frac{\lambda L}{a}\\[4pt] &=\sqrt{\lambda L}+\frac{\lambda L}{\sqrt{\lambda L}}\\[4pt] &=\sqrt{\lambda L}+\sqrt{\lambda L}\\[4pt] &=2\sqrt{\lambda L}\\[4pt] &=\sqrt{4\lambda L}. \end{aligned}$$

Thus the smallest spot is obtained when

$$a = \sqrt{\lambda L}\quad\text{and}\quad b_{\text{min}} = \sqrt{4\lambda L}.$$

These two results coincide exactly with the statements in Option A.

Hence, the correct answer is Option A.