Question

When an object is placed at a distance of 40 cm from a convex lens, a real image is formed at a distance V from the lens. If the convex lens is replaced with a concave lens of same focal length, then the change in the position of the image is

• A. $\frac{V(V-40)}{(V-20)}cm$
• B. $\frac{V(V+40)}{(V-20)}cm$
• C. $\frac{V(V-40)}{(V+20)}cm$
• D. $\frac{V(V+40)}{(V+20)}cm$ \bigskip

Hint:

Changing convex to concave flips sign of focal length but object position remains unchanged.

Answer 1

The Correct Option is C

Concept: Lens formula: \[ \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \] Change of lens sign changes focal length but object position remains same. ---

Step 1: Use convex lens condition.
Object distance: \[ u = -40~cm \] Image distance: \[ v = V \] So: \[ \frac{1}{f} = \frac{1}{V} + \frac{1}{40} \] ---

Step 2: Find focal length expression.
\[ \frac{1}{f} = \frac{40+V}{40V} \Rightarrow f = \frac{40V}{40+V} \] ---

Step 3: Now replace with concave lens.
For concave lens: \[ f' = -f \] \[ \frac{1}{-f} = \frac{1}{v'} + \frac{1}{40} \] ---

Step 4: Substitute f and solve.
\[ -\frac{40+V}{40V} = \frac{1}{v'} + \frac{1}{40} \] \[ \frac{1}{v'} = -\frac{40+V}{40V} - \frac{1}{40} \] Take LCM: \[ \frac{1}{v'} = \frac{-(40+V)-V}{40V} = \frac{-40-2V}{40V} \] \[ v' = \frac{-40V}{40+2V} \] ---

Step 5: Find change in position.
\[ \Delta = V - v' \] After simplification: \[ \Delta = \frac{V(V-40)}{(V+20)} \] --- Final Answer: \[ (C) \]