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**Hint:**In the question, it is given that the light spreads due to diffraction when it passes through the hole. So use the formulas related to diffraction and formulate an equation for the spread of the light on the other side of the wall. Differentiate this equation with respect to the radius of the pinhole to compute the minimum value of the spread.

**Complete step by step answer:**

It is given that the spread of the spot on the wall on the opposite side of the camera is given as the sum of its geometrical spread and the spread due to diffraction. The geometrical spread is the spread due to the hole, so it will have the value of the size of the radius. That is

Geometrical spread $ = a$

Spread due to diffraction$ = \dfrac{{\lambda L}}{a}$

Therefore, the total spread denoted by $b$ will be

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

Taking derivative on both sides with respect to the radius of the hole, we get

$\dfrac{{db}}{{da}} = 1 - \dfrac{{\lambda L}}{{{a^2}}}$

Using the concept of extrema to find the minimum value of the spread, we get

$\dfrac{{db}}{{da}} = 0$

By substituting the value of $\dfrac{{db}}{{da}}$ , we get

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

$ \Rightarrow \dfrac{{\lambda L}}{{{a^2}}} = 1$

Therefore, we get the value of the radius of the pinhole to be

$a = \sqrt {\lambda L} $

Substituting this value of the radius of the pinhole into the equation to find the total spread, we get the minimum value of the size of the spread to be

${b_{\min }} = \sqrt {\lambda L} + \dfrac{{\lambda L}}{{\sqrt {\lambda L} }}$

$ \Rightarrow {b_{\min }} = 2\sqrt {\lambda L} $

If we take $2$ to the inside of the square root, we get the value of the minimum size of spread to be

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

**Hence, we can conclude that option (C) is the correct option.**

**Note:**

Note that we have ignored the negative root of the value for the radius of the pinhole camera because negative length does not have any meaning. Monochromatic light gets diffracted when it passes through a slit of width size that is comparable to the wavelength of the light. The pattern formed is consecutive dark and bright fringes.

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