Ray Optics: Numerical Problems Set (5)

Using $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ with $u = -20$, $f = +15$:

Image $I_1$ forms $60$ cm to the right of the lens.

Mirror is $50$ cm to the right of lens. $I_1$ is $60$ cm to the right of lens, i.e. $10$ cm behind the mirror. So $I_1$ acts as a virtual object for the mirror at $u = +10$ cm (object on the back side).

Wait — sign convention for mirrors: for a concave mirror with light traveling from left to right, the pole is at origin, object distance is negative if object is in front (left). $I_1$ is behind the mirror, so $u = +10$ cm (virtual object).

Mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$, with $f = -10$ (concave), $u = +10$:

Image $I_2$ forms $5$ cm in front of the mirror, i.e. $45$ cm to the right of the lens.

Light now travels right-to-left. Object $I_2$ is at $45$ cm to the right of lens — for the reversed direction, the object is $45$ cm “in front” of the lens. Use lens formula again with $u = -45$:

Image forms $22.5$ cm to the left of the lens (on the same side as the original object). Real and inverted relative to $I_2$.

Final answer: image at $22.5$ cm on the object side of the lens, real.

Why This Works

Multi-element systems work step-by-step: image of one element becomes the object of the next. Sign conventions are the only thing that trips students up — pick “light travels left to right” as positive and stick to it, flipping when light reverses.

Alternative Method

Use the equivalent mirror concept: lens-mirror-lens combination acts as an equivalent mirror with focal length given by $\frac{1}{F} = \frac{2}{f_l} + \frac{1}{f_m}$ when distances permit. Here the geometry does not allow direct simplification, so the step-by-step method is cleaner.