Solution to Question 12
1. Problem Analysis & Setup
We are given a point light source at a distance $L$ from a screen. A converging lens of focal length $f$ and aperture radius $R$ is placed between them. The lens is moved from the source towards the screen.
Let $x$ be the distance of the lens from the source. The distance from the lens to the screen is therefore $L – x$.
Using the lens formula, for an object at distance $u = -x$, the image distance $v$ is given by: $$ \frac{1}{v} – \frac{1}{u} = \frac{1}{f} \implies \frac{1}{v} – \frac{1}{-x} = \frac{1}{f} $$ $$ \frac{1}{v} = \frac{1}{f} – \frac{1}{x} = \frac{x – f}{xf} \implies v = \frac{xf}{x – f} $$
2. Calculating the Spot Radius
The condition $f > L/4$ implies that for a fixed distance $L$, no real image can be formed on the screen (the minimum distance for a real image is $4f$). Thus, the image is formed either behind the screen or is virtual. In either case, the rays intercept the screen to form a circular spot.
Using similar triangles formed by the image formation cone: $$ \frac{r}{R} = \frac{|v – (L – x)|}{|v|} $$ Given the constraints and the nature of the function, this simplifies to: $$ r = R \left[ \frac{x^2 – Lx + Lf}{xf} \right] = \frac{R}{f} \left( x – L + \frac{Lf}{x} \right) $$
3. Minimizing the Spot Size
To find the minimum spot radius, we differentiate $r$ with respect to $x$ and set it to zero: $$ \frac{dr}{dx} = \frac{R}{f} \left( 1 – \frac{Lf}{x^2} \right) = 0 $$ $$ 1 = \frac{Lf}{x^2} \implies x^2 = Lf \implies x = \sqrt{Lf} $$
Now, we substitute $x = \sqrt{Lf}$ back into the expression for $r$: $$ r_{\text{min}} = \frac{R}{f} \left( \sqrt{Lf} – L + \frac{Lf}{\sqrt{Lf}} \right) $$ $$ r_{\text{min}} = \frac{R}{f} \left( \sqrt{Lf} – L + \sqrt{Lf} \right) $$ $$ r_{\text{min}} = \frac{R}{f} (2\sqrt{Lf} – L) $$