A 1.0 cm tall object is 110 cm from a screen, on which is projected a well-focus

Question

A 1.0 cm tall object is 110 cm from a screen, on which is projected a well-focused, 2.0-cm-tall image, from two lenses. One of the lenses has a focal length of -20 cm, and is 20 cm in front of the object.

a) Where is the second lens, and what is its focal length? Remember: no credit unless I can see how you got the answer!

b) Draw a ray diagram TO SCALE, showing both lenses, the object, the image, and enough rays that I can tell you know what you are doing. Show the object distance and both image distances.

Explanation / Answer

we have the object that is at 20 cm from the first diverging lens.

the object distance "p" = 20 cm

The focal lenght = F = - 20 cm

then, we can find where the virtual image produced will be :

1 / 20 + 1 / q = - 1 / 20

q = -10 cm

magnification of the lens = - (-10) / 20 = 0.5 = m1

The object is at 110 cm from a screen, the diverging lens is at 90 cm from the lens, then, the virtual image produced by the lens will be at 100 cm from the screen.

The final image must be at 100 cm from the virtual image, and the final image must be 2 cm tall.

Let's set : x = distance from the virtual objecto to the second lens

y = distance from the virtual object to the second lens

the magnification of the second lens = m2 = - y / x

m2*m1 = -y / 2x >>> total magnification

the final image is twice tall than the first one.

m2*m1 = - 2 = - y / 2x

4x = y

and also : y + x = 100

x = 20 ; y = 80

so then : the focal lenght of the second lens :

1 / 20 + 1 / 80 = 1 / f

f = 16 cm

The second lens must be a converging lens

f = 16 cm

and the location : 80 cm to the right of the screen, and 10 cm from the first diverging lens.

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