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
Less than \(\frac{x}{4}\)
\(\frac{x}{4}\) but less than \(\frac{x}{2}\)
Solution:
For an image of equal size, the magnification is \(m = -1\), which means the image distance \(v = x\) is equal to the object distance \(u = x\). Using the lens formula: \(\frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{2}{x} ⇒ f = \frac{x}{2}\).
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