1. A light source is located at the 0.00 cm point on an optical bench. A converg

ID: 2103291 • Letter: 1

Question

1. A light source is located at the 0.00 cm point on an optical bench. A converging lens with a

focal length magnitude of 8.00 cm is located at the 20.0 cm mark on the optical bench. Next a

diverging lens with a focal length magnitude of 10.00 cm is located at the 28.0 cm mark on the optical

bench.

a) Where on the optical bench should a viewing screen be placed in order to see the real image

formed by this lens combination?

b) Assuming that the light source, or object that is being imaged, has a height of 2.0 cm, what is

the height of its image as viewed on the screen?

f1 = 8 cm

u1 = (0 - 20) = -20 cm

Now, 1 / v - 1 / u = 1 /f

So, 1 / v1 - 1 / -20 = 1 / 8

So, v1 = 13.33 cm.....i.e...at 13.33 cm to the right of the converging lens...

So, its absolute position on the bench = 20 + 13.33 = 33.33 cm

f2 = -10 cm

u2 = -(33.33 - 28) = -5.33 cm

So, 1 / v2 - 1 / u2 - 1 / f2

So, 1 / v2 - 1 / -5.33 = 1 / -10

So, v2 = -3.478 cm

i.e the absolute position = 28 + 3.478 = 31.478 cm

Answer = 31.478 cm

hi / ho = v2 * v1 / u2 * u1 = -3.478 * 13.33 / (-5.33 * -20) = -0.435

So, hi = -0.435 * 2 cm = -0.87 cm

Try entering 0.87 cm instead...it might ask for positiove value only!!!

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