An object is mounted on a wall. Its image of equal size is to be obtained on a parallel wall with the help of

Question

An object is mounted on a wall. Its image of equal size is to be obtained on a parallel wall with the help of a convex lens placed between these walls. The lens is kept at distance x in front of the second wall. The required focal length of the lens will be:

Accepted Answer

1. **Identify Given Conditions:** 
   - A real image is formed on a parallel wall (screen), so the image is real and inverted.
   - The image is of **equal size** to the object, which implies the magnification magnitude $|m| = 1$.
   - The lens is placed at a distance $x$ from the second wall (where the image is formed). Thus, the image distance $v = +x$ (using sign convention where light travels L to R).

2. **Apply Lens Concept:** 
   - For a convex lens, a real image of the same size as the object is formed only when the object is placed at $2f$ and the image is formed at $2f$ on the other side.
   - Therefore, $v = 2f$.

3. **Calculation:** 
   - Since we established $v = x$, we have $x = 2f$.
   - Solving for focal length: $f = x/2$.

4. **Alternative Method (Lens Formula):**
   - Magnification $m = v/u = -1$ (for real, inverted, equal size image).
   - So, $v = -u$. Given $v = x$, then $u = -x$.
   - Lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
   - $\frac{1}{f} = \frac{1}{x} - \frac{1}{-x} = \frac{1}{x} + \frac{1}{x} = \frac{2}{x}$
   - $f = x/2$.

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