Ray Optics: Tricky Questions Solved (9)

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Question

A convex lens of focal length f=20f = 20 cm and a concave mirror of focal length ∣fm∣=15|f_m| = 15 cm are placed coaxially 5050 cm apart. An object is placed 3030 cm in front of the lens (on the side opposite the mirror). Light travels through the lens, reflects off the mirror, and passes through the lens again. Find the position and nature of the final image.

Solution — Step by Step

Using sign convention (distances measured from optical centre, light travels left to right): u=−30u = -30 cm, f=+20f = +20 cm.

1v−1u=1f⇒1v=120−130=160\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} = \frac{1}{20} - \frac{1}{30} = \frac{1}{60}

So v1=+60v_1 = +60 cm. The image forms 60 cm to the right of the lens — but the mirror is only 50 cm away, so this image is behind the mirror, acting as a virtual object for the mirror.

For the mirror: object distance um=+(60−50)=+10u_m = +(60 - 50) = +10 cm (virtual object, on the side from which light comes). Mirror focal length fm=−15f_m = -15 cm (concave, sign convention with light incident on mirror).

1vm+1um=1fm⇒1vm=−115−110=−2+330=−16\frac{1}{v_m} + \frac{1}{u_m} = \frac{1}{f_m} \Rightarrow \frac{1}{v_m} = -\frac{1}{15} - \frac{1}{10} = -\frac{2 + 3}{30} = -\frac{1}{6}

vm=−6v_m = -6 cm. Image forms 6 cm in front of mirror.

Reset sign convention so light moves leftward. Object for the lens is at 50−6=4450 - 6 = 44 cm to the right of the lens.

In standard convention with light direction reversed: u=−44u = -44 cm, f=+20f = +20 cm.

1v=120−144=44−20880=24880=3110\frac{1}{v} = \frac{1}{20} - \frac{1}{44} = \frac{44 - 20}{880} = \frac{24}{880} = \frac{3}{110}

v=110/3≈36.67v = 110/3 \approx 36.67 cm to the left of the lens.

The final image is real, formed approximately 36.6736.67 cm to the left of the lens (on the object side). The system magnification is the product of the three magnifications and turns out negative, so the image is inverted relative to the object.

Why This Works

In multi-element problems, we treat each surface in sequence — image of one becomes object of the next. The hard part is sign conventions: for the mirror leg, light direction matters; for the second lens pass, “light now travels in the opposite direction,” so we flip the sign frame again.

Lens-mirror combinations always have three steps: lens, mirror, lens. Skipping the third step is the most common JEE Main mistake.

Alternative Method

Use the equivalent mirror trick: a thin lens placed in front of a mirror behaves like a mirror with effective focal length 1feq=2fl+1fm\frac{1}{f_{\text{eq}}} = \frac{2}{f_l} + \frac{1}{f_m} — but this only works when lens and mirror are in contact. With 50 cm separation, the sequential approach is necessary.

Common Mistake

Students forget the third step (refraction through the lens on the way back). They stop after the mirror reflection and report that image as “final.” Light always re-enters the lens — every photon that came in must come back through the same elements unless explicitly stated otherwise.

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