Physics Solution – Question 25

Solution to Question 25

1. Velocity Relationships

Let the source velocity be $V_S$ and image velocity be $V_I$. The components of velocity are related by the longitudinal magnification ($m_L$) and transverse magnification ($m$).

At the moment the source crosses the optical axis ($y=0$), the transverse velocity relation is: $$ (V_I)_y = m (V_S)_y $$ The longitudinal velocity relation is: $$ (V_I)_x = m^2 (V_S)_x $$

2. Calculating Magnification from Angles

The angles $\alpha$ and $\beta$ are defined by the ratio of velocity components: $$ \tan\alpha = \frac{(V_S)_y}{(V_S)_x} \quad \text{and} \quad \tan\beta = \frac{(V_I)_y}{(V_I)_x} $$ Substituting the magnification relations into $\tan\beta$:

$$ \tan\beta = \frac{m (V_S)_y}{m^2 (V_S)_x} = \frac{1}{m} \left( \frac{(V_S)_y}{(V_S)_x} \right) $$ $$ \tan\beta = \frac{1}{m} \tan\alpha $$ $$ \Rightarrow m = \frac{\tan\alpha}{\tan\beta} $$

Note: Depending on whether the image is real or virtual, the sign of $m$ can be positive or negative. The ratio of tangents gives the magnitude relation.

3. Finding the Distance

We use the magnification formula in terms of focal length $f$ and object distance $u$: $$ m = \frac{f}{f + u} $$ Solving for $u$:

$$ f + u = \frac{f}{m} $$ $$ u = \frac{f}{m} – f = f\left( \frac{1}{m} – 1 \right) $$

Substituting $m = \frac{\tan\alpha}{\tan\beta}$:

$$ u = f \left( \frac{\tan\beta}{\tan\alpha} – 1 \right) $$

The distance from the lens is $|u|$. Considering both real and virtual image possibilities (where $m$ could be negative relative to the slope ratio), the general solution for the distance is:

Distance $= f\left( 1 \pm \frac{\tan\beta}{\tan\alpha} \right)$