A converging lens, which has a focal length equal to 8.4 cm, is separated by 33.

Question

A converging lens, which has a focal length equal to 8.4 cm, is separated by 33.4 cm from a second lens. The second lens is a diverging lens that has a focal length equal to -16.6 cm. An object is 16.8 cm to the left of the first lens.

(a) Find the position of the final image using both a ray diagram and the thin-lens equation.
____cm to the right of the object
(b) Is the final image real or virtual?

virtual

real    

Is the final image upright or inverted?

upright

inverted    

(c) What is the overall lateral magnification?

Explanation / Answer

Given,

f1 = 8.4 cm ; D = 33.4 cm ; f2 = -16.6 cm ; o1 = 16.8 cm

(a) we need to locate the final image. Let this image distance be i2.

we know from lens equation that,

1/f = 1/i + 1/o

i = o x f / (o - f )

i1 = o1 x f1 / ( o1 - f1)

i1 = 16.8 x 8.4 / (16.8 - 8.4) = 16.8 cm

Now the object distance from the seconf lens will be:

o2 = D - i1 = 33.4 - 16.8 = 16.6 cm

again from lens equation,

i2 = o2 x f2 / (o2 - f2)

i2 = 16.6 x (-16.6) / (16.6 + 16.6) = - 8.3 cm

Hence, i2 = -8.3 cm.

(b)The image is virtual, since the final image distance is negative.

The final image is inverted as the overall magnification is negative(calculated in next part)

(b)we know that,

m1 = -i1/o1 = -16.8/16.8 = -1

m2 = -i2/o2 = -(-8.3)/16.6 = -0.5

M = m1 x m2 = 1 x -0.5 = -0.5

Hence, M = -0.5.

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