HardPhysicsHigh School (11–12)OpticsNumericalConceptual

A small lamp has a bright filament 3.0 cm tall…

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

Two thin lenses on one horizontal principal axis, drawn to scale with a 10 cm scale bar and the heights drawn about twice as large. A blue upright object arrow stands 15 cm to the left of lens L1, which is a convex lens with focal length 10 cm. A second lens L2 is 25 cm to the right of L1. The focal points of L1 are marked at 10 cm on each side of it. The focal points of L2, a concave lens with focal length of magnitude 20 cm, are marked in grey at 20 cm on each side of it. Grey arrows show the distances 15 cm and 25 cm. No rays are drawn.

A small lamp has a bright filament $3.0\ \text{cm}$ tall. It stands $15\ \text{cm}$ in front of a thin convex lens $L_1$ of focal length $10\ \text{cm}$ . A thin concave lens $L_2$ of focal length $20\ \text{cm}$ is placed on the same axis, $25\ \text{cm}$ behind $L_1$ , as the picture shows. A student wants to know where a screen must be placed to catch a sharp picture of the filament. Find the distance of the first image from $L_1$ . Find the distance of the final image from $L_2$ , and say whether it is real. Find the total magnification with its sign. Find the height of the final image with its sign. Take the focal length of the concave lens as negative, and use the direct method one lens at a time.

Need a reminder? The key ideaShow

The power is $P = \dfrac{1}{f}$ , with $f$ in metre and $P$ in dioptre. Lenses in contact: $P = P_1 + P_2$ . Lenses a distance $d$ apart: $P = P_1 + P_2 - dP_1P_2$ , with $d$ in metre. The back focal distance is $F\left(1 - \dfrac{d}{f_1}\right)$ . For two lenses one after the other, the image of the first is the object of the second, and $m = m_1m_2$ .

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.

Given. $h = 3.0\ \text{cm}$ , $u_1 = -15\ \text{cm}$ , $f_1 = +10\ \text{cm}$ , $f_2 = -20\ \text{cm}$ and $d = 25\ \text{cm}$ .

Plan. The image of $L_1$ is the object of $L_2$ . Work out each lens in turn.

Step 1. Lens 1. $v_1 = \dfrac{u_1f_1}{u_1 + f_1} = \dfrac{(-15)(10)}{-15 + 10}\ \text{cm} = 30\ \text{cm}$ and $m_1 = \dfrac{30}{-15} = -2.0$ .

Step 2. The object of lens 2. The first image is $30\ \text{cm}$ from $L_1$ . This is $5\ \text{cm}$ beyond $L_2$ , so $u_2 = 30\ \text{cm} - 25\ \text{cm} = +5\ \text{cm}$ . The object is virtual: the light from $L_1$ would meet $5\ \text{cm}$ behind $L_2$ .

Step 3. Lens 2. $\dfrac{1}{v_2} = \dfrac{1}{u_2} + \dfrac{1}{f_2} = \dfrac{1}{5\ \text{cm}} - \dfrac{1}{20\ \text{cm}} = \dfrac{3}{20\ \text{cm}}$ . So $v_2 = +6.67\ \text{cm}$ and $m_2 = \dfrac{6.67}{5} = +1.33$ .

Step 4. Totals. $m = m_1m_2 = (-2.0)(1.33) = -2.67$ . The height is $-2.67 \times 3.0\ \text{cm} = -8.0\ \text{cm}$ .

Answer: $30\ \text{cm}$ , $6.67\ \text{cm}$ behind $L_2$ , $-2.67$ and $-8.0\ \text{cm}$ . The final image is real, since $v_2$ is positive, so a screen $6.67\ \text{cm}$ behind $L_2$ shows it, inverted.

Watch out. The first image lies beyond $L_2$ , so $u_2$ is $+5\ \text{cm}$ , not $-5\ \text{cm}$ . With the wrong sign you would get $\dfrac{1}{v_2} = -\dfrac{1}{5} - \dfrac{1}{20}$ , so $v_2 = -4.0\ \text{cm}$ , a virtual image. The concave lens makes the rays converge less, but they were already heading for a point behind it, so a real image forms.

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