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A thin convex lens of focal length 20 cm and a concave mirror of radius of curvature 20 cm stand on…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A thin convex lens of focal length $20\ \text{cm}$ and a concave mirror of radius of curvature $20\ \text{cm}$ stand on one optical bench, with a common axis. The mirror faces the lens, $50\ \text{cm}$ from it. A small upright object, $2.0\ \text{cm}$ tall, stands on the axis $30\ \text{cm}$ in front of the lens, on the side away from the mirror, as in the figure. Light from the object passes through the lens, reflects from the mirror, and passes back through the lens.

An optical bench with a horizontal axis. On the left an upright object 2.0 cm tall. 30 cm to its right a thin convex lens of focal length 20 cm. 50 cm to the right of the lens a concave mirror of radius of curvature 20 cm, facing the lens.

A student traces the light one element at a time. The first step is easy. The second step is where most people go wrong, because the light reaches the mirror before it can form the first image.

Where would the lens alone form an image of the object? Give its distance from the lens.
Where does the mirror form its image? Give its distance in front of the mirror.
Where is the final image, after the light has passed back through the lens? Give its distance from the lens.
How tall is the final image? Give a positive value if it is upright and a negative value if it is inverted.
The mirror is now moved along the bench so that the final image falls exactly on the object itself. There are two places where this happens. Give the distance of each from the lens.

Say whether the final image in (c) is real or virtual, and how it could be seen.

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. Trace the light element by element: lens, mirror, lens. At each step, the image of one element is the object of the next, even when it lies behind that element.

Step 1. The lens (a). $\dfrac1v = \dfrac1f - \dfrac1u = \dfrac{1}{20\ \text{cm}} - \dfrac{1}{30\ \text{cm}}$ , so $v = 60\ \text{cm}$ , inverted, twice the size. That is $10\ \text{cm}$ behind the mirror.

Step 2. The mirror (b). The beam is converging towards a point $10\ \text{cm}$ behind the mirror: a virtual object. With distances in front of the mirror positive, $u = -10\ \text{cm}$ and $f = +10\ \text{cm}$ :

\displaystyle \frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{10\ \text{cm}} + \frac{1}{10\ \text{cm}}, \quad v = 5.0\ \text{cm}

A real image $5.0\ \text{cm}$ in front of the mirror. Its magnification is $-v/u = +0.5$ .

Step 3. Back through the lens (c). This image is $45\ \text{cm}$ from the lens and is a real object for it: $\dfrac1v = \dfrac{1}{20\ \text{cm}} - \dfrac{1}{45\ \text{cm}}$ , so $v = 36\ \text{cm}$ , on the object's side of the lens, $6\ \text{cm}$ farther from the lens than the object.

Step 4. Size (d). The magnifications are $-2$ , $+0.5$ and $-0.8$ . Their product is $+0.8$ , so the image is upright and $1.60\ \text{cm}$ tall.

Step 5. Back onto the object (e). The light returns along its own path if the lens's image falls on the mirror's centre of curvature (mirror at $60 + 20 = 80\ \text{cm}$ ) or on its pole (mirror at $60\ \text{cm}$ ). At $80\ \text{cm}$ the final image is inverted; at $60\ \text{cm}$ it is upright.

Answer: (a) $60.0\ \text{cm}$ . (b) $5.0\ \text{cm}$ . (c) $36.0\ \text{cm}$ . (d) $1.60\ \text{cm}$ . (e) $60.0\ \text{cm}$ and $80.0\ \text{cm}$ .

Real or virtual. The final light converges to a point in front of the lens, on the object's side, so the image is real. A screen placed there would show it, and an eye placed well beyond it, looking towards the lens, sees it hanging in the air.

Watch out. Do not treat the first image, $60\ \text{cm}$ from the lens, as a real object $10\ \text{cm}$ in front of the mirror. It is behind the mirror, and the light never gets there.

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