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A convex lens of focal length 15 cm has a plane mirror 25 cm behind it, at right angles to the axis…

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

A convex lens of focal length $15\ \text{cm}$ has a plane mirror $25\ \text{cm}$ behind it, at right angles to the axis. A small object stands on the axis $45\ \text{cm}$ in front of the lens, as in the picture. The light goes through the lens, is reflected by the mirror and goes through the lens again. Find where the lens alone would put its image, where the mirror puts its image (measured from the lens), where the final image is, and the total magnification. Use the New Cartesian convention at each step.

A horizontal axis with a convex lens at the middle, focal length 15 cm, with its two foci F marked at 15 cm on each side. A blue object arrow is 45 cm to the left of the lens. A plane mirror, a vertical line with hatching, is 25 cm to the right of the lens. Double-headed arrows mark the distances 45 cm and 25 cm below the axis.

Need a reminder? The key ideaShow

A lens with one silvered face acts as a mirror of power $P = 2P_{lens} + P_{mirror}$ . A plane face has $P_{mirror} = 0$ and a silvered curved face has $P_{mirror} = \dfrac{2}{R}$ . For the mirror, $\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{F}$ with $u$ and $v$ positive in front. A lens cut across keeps $\dfrac{n - 1}{R}$ in each piece.

The worked solution

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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.

Given. $f = 15\ \text{cm}$ , the mirror is $25\ \text{cm}$ from the lens and $u = -45\ \text{cm}$ .

Plan. First pass through the lens, then the mirror, then the lens again.

Step 1. First pass. $\dfrac{1}{v_1} = \dfrac{1}{f} + \dfrac{1}{u} = \dfrac{1}{15\ \text{cm}} - \dfrac{1}{45\ \text{cm}} = \dfrac{2}{45\ \text{cm}}$ , so $v_1 = 22.5\ \text{cm}$ behind the lens. The magnification is $m_1 = \dfrac{22.5\ \text{cm}}{-45\ \text{cm}} = -0.500$ .

Step 2. The mirror. The point at $22.5\ \text{cm}$ is $2.5\ \text{cm}$ in front of the mirror. The light reaches the point before the mirror. The mirror puts the image $2.5\ \text{cm}$ behind it, which is $25\ \text{cm} + 2.5\ \text{cm} = 27.5\ \text{cm}$ from the lens. The magnification of the mirror is $+1$ .

Step 3. Second pass. The light now travels to the left, diverging from the point $27.5\ \text{cm}$ from the lens. This is a real object, so $u_2 = -27.5\ \text{cm}$ . We want $v_2$ , so $\dfrac{1}{v_2} = \dfrac{1}{15\ \text{cm}} - \dfrac{1}{27.5\ \text{cm}}$ and $v_2 = 33.0\ \text{cm}$ in front of the lens. The magnification is $m_2 = \dfrac{33.0\ \text{cm}}{-27.5\ \text{cm}} = -1.20$ .

Step 4. The total. $m = (-0.500)(+1)(-1.20) = +0.600$ . The final image is upright and $0.600$ times the object.

Answer: $22.5\ \text{cm}$ , $27.5\ \text{cm}$ , $33.0\ \text{cm}$ and $0.600$ .

Check. The two inversions of the lens cancel, so the image is upright, which agrees with the sign of the product.

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