An object 1.20 cm high is placed 6.00 cm in front of converging lens (lens#1) wi

Question

An object 1.20 cm high is placed 6.00 cm in front of converging lens (lens#1) with a focal length magnitude of 12.0 cm. A second converging lens (lens#2) with a focal length of 6.00 cm is placed 12.0 cm behind the original lens (a) Where is the final image located relative to lens #2? (b) what is the final image height & (c) what are the final image characteristics?

Hint: the image from the lens is the object for the mirror & the final image characteristics are relative to the original object.

Explanation / Answer

By using the lens formula :

1/V+1/u=1/f

1/V+1/6=1/12

V= -12 cm

Thus image is infront of converging lens at distance 12 cm.

Hi= -v*Ho/u

=12*1.2/6

=2.4 cm

Distance of image from converging lens = 12+12 =24 cm

Again using the lens formula for second converging lens:

1/V' + 1/24 = 1/6

V' = 8 cm

(a)

Thus image is located behing the second lens at distance 8 cm .

(b)

Hi'=-8*2.4/24

= - 0.8 cm

Thus, the image height is 0.8 cm

(c)

characteristics of final image :

Real

Inverted

Smaller in size .

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