A 1.20-cm-tall object is 50.0 cm to the left of a converging lens of focal lengt

Question

A 1.20-cm-tall object is 50.0 cm to the left of a converging lens of focal length 40.0 cm. A second converging lens, this one having a focal length of 60.0 cm, is located 300.0 cm to the right of the rst lens along the same optic axis. Find the location and height of the images by the first and second lenses, with the second image being the final image.

Repeat using the same lenses except for the following changes: (a) The second lens is a diverging lens having a focal length of magnitude 60.0 cm. (b) The rst lens is a diverging lens having a focal length of magnitude 40.0 cm. (c) Both lenses are diverging lenses having focal lengths of the same magnitudes as in Problem 34.39.

I understand the initial problem and (a), but I'm getting confused with (b) in that I'm not sure how to interpret the signs from my math compared to my diagram. Could someone go through the math and the corresponding diagram?

Explanation / Answer

A) 1/u+1/v=1/f.
u=50, f=40
1/v=1/40-1/50
200/v=5-4=1
v=+200cm (200cm to the right of the 1st lens).
(Note: this distance seems quite far, but the object is quite near the focal point; if it had been at the focal point the image would have been formed at +infinity.)
magnification=v/u=200/50=4
so image height is 4 x 1.20 = +4.80 cm (erect).

B) 1/u+1/v=1/f.
u=300-200=100, f=-60.
1/v=1/-60-1/100
300/v=-5-3=-8
v=-300/8=-37.5cm (to the left of the 2nd lens, so the image is virtual).
magnification=v/u=-37.5/100=-0.375
so image height hv = -0.375 x 4.80 = -1.80 cm

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