An object is placed 15.0 cm from a first converging lens of focal length 10.0 cm

Question

An object is placed 15.0 cm from a first converging lens of focal length 10.0 cm. A second converging lens with focal length 5.00 cm is placed 40.0 cm to the right of the first converging lens.

(a) Use graph paper to construct a ray diagram that you will use to determine the location of the image and its magnification. Do this by producing an image from the first lens as if the second lens did not exist. Then use this image as an object for the second lens. What is the final image position, based on your ray tracing? Also, from the same ray tracing diagram determine whether the final image is upright or inverted.

(b) Use the standard thin-lens equation to determine the solution without using ray tracing. Again, start by finding the image from the first lens and using that image as the object for the second lens. Find the final magnification and compare your answer to what you had obtained using part (a).

(c) Instead of placing the second lens 40 cm to the right of the first you now put it 10 cm to the right of the first converging lens. Show that this means that the first image is now a virtual object for the second and use the thin-lens equation approach to find the final position of the resulting image (only use the thin-lens equation here and do not bother with the ray tracing tool). Is the final image real or virtual?

(d) Obtain also the final magnification for the position described in part (c). Is the final image inverted or upright?

Explanation / Answer

(a) first image's position:
1/f1 = 1/q1 + 1/p1
1/q1 = 1/46 - 1/17 = -29/782
q1 = -782/29 cm = -27 cm (virtual image)

(b) 27 cm + 10.3 cm = 37.3 cm

(c) p2 = 37.3 cm

(d) second image's position:
1/f2 = 1/q2 + 1/p2
1/q2 = 1/50 - 1/37.3 = -127/18650
q2 = -18650/127 cm = -146.9 cm (virtual image)

(e) M1 = -q1/p1 = 27/17 = 1.6

(f) M2 = -q2/p2 = 146.9/37.3 = 3.9

(g) M = M1 * M2 = 25/4

(h) virtual and upright image

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