An object is located at a distance of 4 cm from a thin converging lens with foca

Question

An object is located at a distance of 4 cm from a thin converging lens with focal length of 2 cm. A diverging lens is located 3 cm from the converging lens and 7 cm from the object. The diverging lens has a focal length of -2 cm. Note: To handle a multiple lens system, we treat them independently. We first find the image created by the first lens. We then use the image from the first lens to act as the object for the second lens.

Use the thin lens equation (1/f = 1/s' + 1/s) to predict the following:

(a) Location of the image?

(b) Magnification of the image (including inverted versus non-inverted)?

(c) Real or virtual?

Explanation / Answer

do = 4 cm
f = 2 cm

Now,
1/di + 1/do = 1/f
1/di + 1/4 = 1/2
di = 4 cm

M1 = -di/do = - (4/4) = -1

Now image from the first lens to act as the object for the second lens.

do = - 1 cm
f = - 2 cm

1/di + 1/do = 1/f
1/di - 1/1 = -1/2
di =  2 cm

M2 = -di/do
M2 = - 2/(-1) = 2

(a)
Location of the image, 2cm from the diverging lens.

(b)
M = M1*M2
M = -1 * 2
M = -2
Magnification of the image, M = -2
Image is Inverted.

(c)
Image is Virtual.

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