Problem 66

  A converging lens (f = 12.0 cm) is 28.0 cm to the left of a diverging lens (f = -14.0 cm). An object is located 6.00 cm to the left of the converging lens. Draw an accurate ray diagram and from it find (a) the final image distance, measured from the diverging lens, and the overall magnification. (b) Confirm your answers to part (a) by using the thin-lens and magnification equations.

SOLUTION:
Again, I'll try to explain what steps you should take in drawing the ray diagram. First, consider only the nearest lens, the converging one in this case. Draw a ray from the top of the object through the center of the lens. Draw another ray paraxially, which bends toward the focus on the other side (continue the line backwards, also). The rays meet behind the object. Using the lens equation, we find that

$$\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o}$$

$$\frac{1}{d_i} = \frac{1}{12 cm} - \frac{1}{6 cm}$$

$$\frac{1}{d_i} = -\frac{1}{12 cm}$$

d_i = -12 cm

The image is located 12 cm to the left of the converging lens. It has a magnification

$$m = -\frac{d_i}{d_o} = -\frac{-12 cm}{6 cm} = 2 .$$

Now, we'll use this image as the object for the farther lens, in this case the diverging lens.
Draw a ray from the top of the new object (the old image) through the center of the diverging lens. Also draw a ray from the top of the object paraxially, and continue it backwards through the focal point of the diverging lens. You'll find a little image to the left of the diverging lens but in front of its focus.
More quantitatively, since the diverging lens is 28 cm to the right of the converging lens, the object distance is d_o = 28 cm + 12 cm = 40 cm; this is the distance between the diverging lens and the image formed by the converging lens. Using the lens equation, we get

$$\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o}$$

$$\frac{1}{d_i} = \frac{1}{-14 cm} - \frac{1}{40 cm}$$

d_i = -10.4 cm

That is, the final image is 10.4 cm to the left of the diverging lens. Now, the magnification due to the diverging lens is

$$m = -\frac{-10.4 cm}{40 cm} = 0.26$$

Because the magnification of the converging lens was m = 2, the overall magnification is

m = 2(0.26) = 0.52 .