Question

a compound lens system consists of two converging lenses, one at x = -20.0 cm with focal length f₁ = +10.0 cm, and the other at x = +30.0 cm with focal length f₂ = +8.00 cm (figure 1). an object 1.00 centimeter tall is placed at x = -50.0 cm.
figure
1 of 1
part a
what is the location of the final image produced by the compound lens system? give the x coordinate of the image.
express your answer in centimeters, to three significant figures or as a fraction.
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x = cm
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part b complete previous part(s)
part c complete previous part(s)

Explanation:

Step1: Image from first lens

The object distance for the first lens ($f_1 = 10.0$ cm) is $d_{o1} = -50.0 - (-20.0) = -30.0$ cm (since object is at $x=-50.0$ and lens at $x=-20.0$). Using the lens formula $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$, for first lens:
$\frac{1}{d_{i1}} = \frac{1}{f_1} - \frac{1}{d_{o1}} = \frac{1}{10.0} - \frac{1}{-30.0} = \frac{3 + 1}{30.0} = \frac{4}{30.0}$
So $d_{i1} = \frac{30.0}{4} = 7.50$ cm. This image is at $x_1 = -20.0 + 7.50 = -12.5$ cm (relative to origin).

Step2: Object for second lens

The distance between lenses is $50.0 - (-20.0) = 70.0$ cm. The object distance for the second lens ($f_2 = 8.00$ cm) is $d_{o2} = 50.0 - (-12.5) = 62.5$ cm (since image from first lens is at $x=-12.5$, lens at $x=50.0$). Wait, no: Wait, the first lens is at $x=-20$, image at $x=-20 + 7.5 = -12.5$ (left of first lens? Wait no, converging lens: if $d_{o1}$ is negative (object on left), $d_{i1}$ positive means image on right of first lens. Wait, correction: $d_{o1}$ is object distance from first lens: object is at $x=-50$, lens at $x=-20$, so $d_{o1} = -20 - (-50) = 30.0$ cm (positive, object on left of lens). Oh! I messed up sign convention. Lens formula: $d_o$ is positive when object is on left (real object). So $d_{o1} = 30.0$ cm (object at $x=-50$, lens at $x=-20$: distance is $30$ cm left of lens? No, object is left of lens, so $d_{o1} = 30.0$ cm (positive). Then $\frac{1}{d_{i1}} = \frac{1}{10.0} - \frac{1}{30.0} = \frac{2}{30.0}$, so $d_{i1} = 15.0$ cm. So image from first lens is at $x = -20.0 + 15.0 = -5.0$ cm? Wait no, lens at $x=-20$, image 15 cm to the right: $x = -20 + 15 = -5.0$ cm. Now, distance to second lens (at $x=50.0$) is $50.0 - (-5.0) = 55.0$ cm. So $d_{o2} = 55.0$ cm (object on left of second lens, positive).

Step3: Image from second lens

Using lens formula for second lens: $\frac{1}{d_{i2}} = \frac{1}{f_2} - \frac{1}{d_{o2}} = \frac{1}{8.00} - \frac{1}{55.0} = \frac{55 - 8}{440} = \frac{47}{440}$
Thus, $d_{i2} = \frac{440}{47} \approx 9.36$ cm. So the final image is at $x = 50.0 + 9.36 = 59.4$ cm? Wait, no, wait: first lens image at $x=-5.0$ (left of second lens? No, second lens is at $x=50.0$, so $x=-5.0$ is left of second lens, so $d_{o2} = 50.0 - (-5.0) = 55.0$ cm (correct, object on left of second lens, positive). Then image from second lens is $d_{i2}$ to the right of second lens, so $x = 50.0 + 9.36 = 59.4$ cm? Wait, but let's recalculate $\frac{1}{d_{i2}} = \frac{1}{8} - \frac{1}{55} = \frac{55 - 8}{440} = \frac{47}{440} \approx 0.1068$, so $d_{i2} \approx 9.36$ cm. So final x-coordinate is $50 + 9.36 = 59.4$ cm? Wait, but maybe I messed up first lens. Let's redo first lens:

First lens: $f_1 = 10$ cm, object at $x=-50$, lens at $x=-20$. So object distance $d_{o1} = -20 - (-50) = 30$ cm (positive, object on left). Lens formula: $\frac{1}{10} = \frac{1}{30} + \frac{1}{d_{i1}} \implies \frac{1}{d_{i1}} = \frac{3 - 1}{30} = \frac{2}{30} \implies d_{i1} = 15$ cm. So image at $x = -20 + 15 = -5$ cm (right of first lens, left of second lens, since second lens is at $x=50$). Now, distance from this image to second lens: $50 - (-5) = 55$ cm (object distance for second lens, $d_{o2} = 55$ cm, positive). Second lens: $f_2 = 8$ cm. $\frac{1}{8} = \frac{1}{55} + \frac{1}{d_{i2}} \implies \frac{1}{d_{i2}} = \frac{55 - 8}{440} = \frac{47}{440} \implies d_{i2} = \frac{440}{47} \approx 9.36$ cm. So final image at $x = 50 + 9.36 = 59.4$ cm? Wait, but let's check again. Wait, maybe the first image is virtual? No, converging lens, object outside focal length (30 cm > 10 cm), so real image. Wait, but 30 cm object distance, focal length 10 cm: image at 15 cm (correct, real image). Then that image is at $x = -20 + 15 = -5$ cm (so 5 cm left of origin, 55 cm left of second lens? Wait no, second lens is at $x=50$, so distance from $x=-5$ to $x=50$ is 55 cm (object on left of second lens, so $d_{o2} = 55$ cm, correct). Then second lens: $f=8$, $d_o=55$, so $d_i = \frac{1}{\frac{1}{8} - \frac{1}{55}} = \frac{440}{55 - 8} = \frac{440}{47} \approx 9.36$ cm. So final x is $50 + 9.36 = 59.4$ cm. Wait, but maybe I made a mistake in sign for first lens. Let's use standard sign convention: for a lens, object distance $d_o$ is positive if object is on the left (real object), image distance $d_i$ positive if image on right (real image). So first lens: $d_o = 30$ cm, $f=10$ cm. $\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} = \frac{1}{10} - \frac{1}{30} = \frac{2}{30} \implies d_i = 15$ cm (right of first lens, so at $x = -20 + 15 = -5$ cm). Then second lens: $d_o = 50 - (-5) = 55$ cm (left of second lens, positive), $f=8$ cm. $\frac{1}{d_i} = \frac{1}{8} - \frac{1}{55} = \frac{55 - 8}{440} = \frac{47}{440} \implies d_i = \frac{440}{47} \approx 9.36$ cm (right of second lens, so at $x = 50 + 9.36 = 59.4$ cm). So the final x-coordinate is approximately 59.4 cm.

Answer:

$\boxed{59.4}$ (or exact fraction $\frac{440}{47} \approx 9.36$? Wait no, wait: Wait, no, the final x is 50 + 9.36 = 59.4. Wait, let's recalculate $\frac{440}{47} \approx 9.3617$, so 50 + 9.3617 = 59.3617 ≈ 59.4 cm (three significant figures).