An 4.00 cm tall object is positioned 20.0 cm to the left of a diverging lens of

Question

An 4.00 cm tall object is positioned 20.0 cm to the left of a diverging lens of focal length |f| = 32 cm.

a. Where is the image located (how far and to which side of the lens)?

b. How big is the image?

c.   Is the image inverted or upright?

d.      If you wanted to see this image, describe the method you would use to view it.

e.       draw an accurate Ray Diagram.  

f.     consider that an object is placed 36 cm in front of a converging lens of focal length f1= 13 cm, which in turn is 56 cm in front of another converging lens of focal length f2= 16 cm. Draw an accurate ray diagram to find the location of the image.  

g.      Relative to the first lens, where is the image located?

h.      Is it a real or virtual image?

i.        What is the total magnififcation?

Explanation / Answer

Yes you use the same eqn except f is negative

so 1/f = 1/s + 1/s' where s is the object distance and s' is the image distance

therefore 1/s' = 1/f - 1/s = 1/(32) - 1/20 = -0.01875 so s' = - 53.33 cm

SO the image is 53.33 cm in front of the lens

and the height y' = m*y where m = -s'/s

so y = 4.0*53.33/20 = 10.66 cm

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