Three identical convex lenses each of focal length f are placed in a straight line separated by a distance f from each other. An object is located at f/2 in front of the leftmost lens. Then,
Final image will be at f/2 behind the rightmost lens and its magnification will be −1.
For a system of lenses, the image formed by each lens acts as the object for the next, and the total magnification is the product of the individual magnifications. Using the Cartesian sign convention, \(\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}\) and \(m = \dfrac{v}{u}\).
Step 1 — First lens: \(u_1 = -\dfrac{f}{2}\), so \(\dfrac{1}{v_1} = \dfrac{1}{f} - \dfrac{2}{f} = -\dfrac{1}{f}\), giving \(v_1 = -f\) (virtual image, f to the left of the first lens) and \(m_1 = \dfrac{-f}{-f/2} = 2\).
Step 2 — Second lens is f to the right of the first, so the first image is 2f to its left: \(u_2 = -2f\), \(\dfrac{1}{v_2} = \dfrac{1}{f} - \dfrac{1}{2f} = \dfrac{1}{2f}\), \(v_2 = 2f\), \(m_2 = \dfrac{2f}{-2f} = -1\).
Step 3 — This image lies 2f right of the second lens, i.e. f to the right of the third lens, so it is a virtual object: \(u_3 = +f\), \(\dfrac{1}{v_3} = \dfrac{1}{f} + \dfrac{1}{f} = \dfrac{2}{f}\), \(v_3 = \dfrac{f}{2}\), \(m_3 = \dfrac{f/2}{f} = \dfrac{1}{2}\).
Step 4 — Total magnification \(m = m_1 m_2 m_3 = 2\times(-1)\times\dfrac{1}{2} = -1\), with the final image f/2 behind the rightmost lens.
A magnification of +1 would mean an erect final image, but the second lens inverts it and no later lens inverts it back. An image distance of f behind the last lens would require the light entering the third lens to be parallel, whereas it is converging towards a point f beyond it.
Final image will be at f/2 behind the rightmost lens and its magnification will be −1. option is correct.
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Important Questions from Optics
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