Physics Solution – Question 24

Solution to Question 24

1. Analyze the Source Coordinates

The point source $S$ moves on a circular path centered at the focus $F$ of the converging lens. From the problem description:

Focal length of lens = $f$.
Let the lens be at the origin $(0, 0)$.
Center of circle (Object Focus $F$) = $(-f, 0)$.
Radius of circle $r = f/2$.

We can parameterize the position of the source $(u, y_S)$ using the angle $\theta$:

$$u = -f + r\cos\theta = -f + \frac{f}{2}\cos\theta = -f\left(1 – \frac{1}{2}\cos\theta\right)$$ $$y_S = r\sin\theta = \frac{f}{2}\sin\theta$$

2. Determine the x-coordinate of the Image ($x$)

Using the thin lens formula $\frac{1}{v} – \frac{1}{u} = \frac{1}{f}$, where $v = x$:

$$ \frac{1}{x} = \frac{1}{f} + \frac{1}{u} $$ $$ \frac{1}{x} = \frac{1}{f} + \frac{1}{-f\left(1 – \frac{1}{2}\cos\theta\right)} $$ $$ \frac{1}{x} = \frac{1}{f} \left( 1 – \frac{1}{1 – \frac{1}{2}\cos\theta} \right) $$ $$ \frac{1}{x} = \frac{1}{f} \left( \frac{1 – \frac{1}{2}\cos\theta – 1}{1 – \frac{1}{2}\cos\theta} \right) = \frac{1}{f} \left( \frac{-\frac{1}{2}\cos\theta}{1 – \frac{1}{2}\cos\theta} \right) $$ $$ \frac{1}{x} = \frac{-\cos\theta}{f(2-\cos\theta)} $$ $$ x = -f \frac{2-\cos\theta}{\cos\theta} = -f \left( \frac{2}{\cos\theta} – 1 \right) $$ $$ x = f \left( 1 – \frac{2}{\cos\theta} \right) = f(1 – 2\sec\theta) $$

3. Determine the y-coordinate of the Image ($y$)

The magnification $m$ is given by $m = \frac{v}{u} = \frac{x}{u}$.

$$ m = \frac{f(1 – 2\sec\theta)}{-f(1 – \frac{1}{2}\cos\theta)} = -\frac{1 – \frac{2}{\cos\theta}}{1 – \frac{\cos\theta}{2}} $$ $$ m = -\frac{\frac{\cos\theta – 2}{\cos\theta}}{\frac{2 – \cos\theta}{2}} = -\frac{\cos\theta – 2}{\cos\theta} \cdot \frac{2}{-(\cos\theta – 2)} $$ $$ m = \frac{2}{\cos\theta} = 2\sec\theta $$

Now, calculating $y$:

$$ y = m \cdot y_S $$ $$ y = (2\sec\theta) \cdot \left( \frac{f}{2}\sin\theta \right) $$ $$ y = f \frac{\sin\theta}{\cos\theta} = f \tan\theta $$

Figure: Locus of the image (Hyperbola) with asymptotes shown in red.

4. Equation of the Locus

To find the locus, we eliminate the parameter $\theta$. From the y-equation: $$ \tan\theta = \frac{y}{f} $$ From the x-equation: $$ \sec\theta = \frac{1 – x/f}{2} = \frac{f-x}{2f} $$ Using the trigonometric identity $\sec^2\theta – \tan^2\theta = 1$:

$$ \left( \frac{f-x}{2f} \right)^2 – \left( \frac{y}{f} \right)^2 = 1 $$ $$ \frac{(x-f)^2}{4f^2} – \frac{y^2}{f^2} = 1 $$

This is the equation of a hyperbola centered at $(f, 0)$.

Center: $(f, 0)$
Vertices: $(3f, 0)$ and $(-f, 0)$
Asymptotes: $\frac{x-f}{2f} = \pm \frac{y}{f} \implies y = \pm \frac{1}{2}(x-f)$. These are shown as the red lines in the diagram.

Coordinates: $x = f\left(1 – \frac{2}{\cos\theta}\right), \quad y = f\tan\theta$
Locus: Hyperbola $\frac{(x-f)^2}{4f^2} – \frac{y^2}{f^2} = 1$