What would be the position of image formed by the lens combination gi…

The correct answer is Option (1) → 10 cm to the right of third lens

For a combination of lenses, the image formed by each lens acts as the object for the next lens. Use the thin lens formula with the Cartesian sign convention (distances measured from the lens, positive to the right, light travelling left to right).

    \( \displaystyle \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \)

First lens (\( f_1 = +10\,\mathrm{cm} \), object at \( u_1 = -20\,\mathrm{cm} \)):

    \( \displaystyle \frac{1}{v_1} = \frac{1}{10} + \frac{1}{-20} = \frac{1}{20} \;\Rightarrow\; v_1 = +20\,\mathrm{cm} \)

The second lens is 10 cm to the right of the first, so this image lies 10 cm to the right of the second lens. It acts as a virtual object for the second lens.

Second lens (\( f_2 = -10\,\mathrm{cm} \), \( u_2 = +10\,\mathrm{cm} \)):

    \( \displaystyle \frac{1}{v_2} = \frac{1}{-10} + \frac{1}{10} = 0 \;\Rightarrow\; v_2 = \infty \)

So the light leaves the second lens as a parallel beam.

Third lens (\( f_3 = +10\,\mathrm{cm} \), \( u_3 = \infty \)):

    \( \displaystyle \frac{1}{v_3} = \frac{1}{10} + 0 \;\Rightarrow\; v_3 = +10\,\mathrm{cm} \)

A parallel beam is focused at the focal point, so the final image is 10 cm to the right of the third lens. The 20 cm gap between the second and third lenses does not matter, because the beam between them is parallel.