Question

A converging lens has a focal length of 7.5 cm.

(a) What are the image distances for objects placed at the following distances from the thin lens? Also, state if each image is real or virtual, upright or inverted, and enlarged or diminished in size.

Object Distance	Image Distance	Real/Virtual	Upright/Inverted	Enlarged/Same Size/Diminished in Size
5.0 cm	cm	---Select--- real virtual	---Select--- upright inverted	---Select--- enlarged same size diminished in size
14.5 cm	cm	---Select--- real virtual	---Select--- upright inverted	---Select--- enlarged same size diminished in size
16.0 cm	cm	---Select--- real virtual	---Select--- upright inverted	---Select--- enlarged same size diminished in size
25.0 cm	cm	---Select--- real virtual	---Select--- upright inverted	---Select--- enlarged same size diminished in size

(b) If the object is 5.00 cm high, what is the height of the image for the object distance of 5.0 cm and 25.0 cm?

Object Distance	Image Height
5.0 cm	cm
25.0 cm	cm

Answer 1

(a)

(1) If object distance is less than its focallength distance, ( 5,0 cm < 7.5 cm )

In the case of object placed infront of converging lens, the image is located behind the object. In this case, the image will be an vitual,   upright image, and enlarged in size

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(2)

In the case of the object is located between 2F and F ( here F is focal length): 15 cm > 14.5 cm > 7.5 cm

In this case, image is real, inverted and enlarged

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(3)

In this case, the object is located beyond 2F

In this case, the image is real, inverted and diminished in size.

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(4)

In this case, the object is located beyond 2F

In this case, the image is real, inverted and diminished in size

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(b)

Use the definition of magnification

M = hi / ho = - u / v

Rearrange for image height hi

hi = ( - u / v )  ho

In order to find height of image, we need image distance ( unknown). To find this image distance, we use lens equation

lens equation

1/f = 1/u + 1/v

Here, focal lenth , f is also not given. so, we couldn't solve for image height with two unknowns.