Ray Optics: PYQ Walkthrough (8)

Using $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$ with $u = -30$ cm, $f = +20$ cm:

$\frac{1}{v} = \frac{1}{20} + \frac{1}{-30} = \frac{3 - 2}{60} = \frac{1}{60}$

$v = +60$ cm. The image is 60 cm to the right of the lens — but the mirror is only 50 cm away, so this image forms 10 cm behind the mirror.

For the mirror, the object is 10 cm behind it (virtual object), so $u_{\text{mirror}} = +10$ cm using mirror sign convention (object on the same side as incident light is negative; here it’s behind the mirror, which is the positive side).

Wait — careful with sign convention. Mirror sign convention: distances measured from the pole, against the direction of incident light are positive. Light travels left-to-right, hits the mirror; the virtual object is to the right of the mirror (behind it), so $u = +10$ cm.

$f_{\text{mirror}} = -10$ cm (concave). Using $\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$:

$\frac{1}{v} = -\frac{1}{10} - \frac{1}{10} = -\frac{2}{10}$

$v = -5$ cm. The image is 5 cm in front of the mirror (on the lens side).

The mirror image is 5 cm in front of the mirror, so $50 - 5 = 45$ cm from the lens, on the right side. For the second pass through the lens, $u = +45$ cm (object on the side of outgoing light from the mirror, which becomes incident light for the lens going leftward).

Using lens formula again:

$\frac{1}{v} = \frac{1}{20} + \frac{1}{45} = \frac{9 + 4}{180} = \frac{13}{180}$

$v \approx 13.85$ cm to the left of the lens.

Final answer: Image forms about 13.85 cm on the left side of the lens (same side as the original object).