Question

Consider a convex lens of focal length $f$. A point object moves towards the lens along its axis between $2f$ and $f$. If the speed of the object is $V_o$, then its image would move with speed $V_i$. Which of the following is correct?

• A. $V_i = V_o$; the image moves away from the lens.
• B. $V_i = -V_o$; the image moves towards the lens.
• C. $V_i > V_o$; the image moves away from the lens.
• D. $V_i < V_o$; the image moves away from the lens.

Hint:

When an object moves between $2f$ and $f$, its real image moves even faster on the opposite side of the lens.

Answer 1

The Correct Option is C

Step 1: Use the lens formula.
$\displaystyle \frac{1}{v} + \frac{1}{u} = \frac{1}{f}$, where $u<0$ for a real object. Differentiating w.r.t. time $t$: $\displaystyle -\frac{1}{u^2}\frac{du}{dt} - \frac{1}{v^2}\frac{dv}{dt} = 0.$

Step 2: Relate object and image velocities.
Let object velocity $V_o = -\dfrac{du}{dt}$ (positive towards the lens), and image velocity $V_i = \dfrac{dv}{dt}$. Then from the differentiated lens equation: $\displaystyle \frac{V_o}{u^2} = \frac{V_i}{v^2}.$ Thus $\displaystyle V_i = V_o \left(\frac{v}{u}\right)^2.$

Step 3: Behaviour between $2f$ and $f$.
When $f<u<2f$, the image lies beyond $2f$ on the other side and $|v| > |u|$. So $\displaystyle \left(\frac{v}{u}\right)^2 > 1 \;\Rightarrow\; V_i > V_o.$

Step 4: Direction.
As the object approaches the lens, the image moves away from the lens (positive $V_i$).

Step 5: Conclusion.
The image moves faster than the object and in the opposite direction: $V_i > V_o$, moving away from the lens ⇒ option (C).