An object is placed 100 cm in front of a diverging lens of focal length -25 cm.

Question

An object is placed 100 cm in front of a diverging lens of focal length -25 cm. A converging lens of focal length 33 cm is placed 30 cm past the first lens. What is the lateral magnification of this system of lenses?

1.0
2.5
-0.4
0.4
-2.5

An object is placed 100 cm in front of a lens of focal length 50 cm. A lens of focal length -20 cm is placed 90 cm beyond the first lens. The final image is located

An object is placed on an optical bench. A lens with a focal length of 12 cm is placed 20 cm from the object. Where should a second lens, with a focal length of 4.0 cm, be placed if a total magnification +6 is to be obtained?

30 cm from the first lens
25 cm from the first lens
20 cm from the first lens
40 cm from the first lens
35 cm from the first lens

A thin converging lens is found to form an image of a distant building 24 cm from the lens. If an object is placed 16 cm from the lens, how far from the object will the image be?

96 cm
48 cm
32 cm
64 cm
72 cm

17 cm past the second lens. 10 cm past the second lens. 10 cm in front of the second lens. 20 cm past the second lens. 20 cm in front of the second lens.

Explanation / Answer

4)image distance,di= 24 cm

object distance,do= 16 cm

1/f = 1/do + 1/di

Since building is very far, do is very large and 1/do = zero compared to 1/di

Thus, 1/f = 1/di and the focal length is 24 cm.

Next use focal length in the problem

1/f = 1/do + 1/di

1/24 = 1/16 + 1/di

2/48 - 3/48 = 1/di

-1/48 = 1/di

di = - 48 cm

The answer is negative because since it is a VIRTUAL image on the same side as the object.

So, 48 cm - 16 cm = 32 cm

Image is 32 cm away from the object. Image is magnified by 3.

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