An illuminated arrow forms a real, inverted image of itself at a distance d = 50

Question

An illuminated arrow forms a real, inverted image of itself at a distance d = 50 cm, measured along the central axis of a lens as drawn below. The image is twice the size of the object:

A) What kind of lens must be used?

B) Where must it be placed?

C) What is its focal length?

Please show your solution step by step with details

An illuminated arrow forms a real, inverted image of itself at a distance d = 50 cm, measured along the central axis of a lens as drawn below. The image is twice the size of the object: A) What kind of lens must be used? B) Where must it be placed? C) What is its focal length? Please show your solution step by step with details

Explanation / Answer

A) A converging (convex) lens must be used, because a diverging (concave) lens forms only a virtual image that is smaller than the object.

B) The magnification is equal to the negative of the image distance divided by the object distance, with the standards signs used for these distances, and is also the ratio of the image height to the object height.

M = - di / do= hi/ ho

Because the image is inverted, the image height hi is negative, while the object height ho is positive, and the ratio hi/ ho = -2

M = - di / do= hi/ ho = -2

Rearrange

di = 2 do

We are given that the total of the image and object distance is 50 cm.

do + di = 50 cm

Use the equation before this

do + 2do = 50 cm

do = (50 cm)/3 =16.7 cm

do = 50 cm - 16.7 cm = 33.3 cm

So the lens must be placed 33.3 cm to the right of the object.

The object distance is 25 cm, and the positive sign indicates it is on the left of the lens.

C) Use the lens equation

1/di +1/do = 1/f

1/(33.3 cm) + 1/(16.7 cm) = 1/f

3/(33.3 cm) = 1/f

f = 11.1 cm

f = 16.7 cm

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