1. Keplerian Telescope An object is placed 120 cm to the left of a converging le

Question

1. Keplerian Telescope

An object is placed 120 cm to the left of a converging lens (f1=40 cm). This lens is 90 cm to the left of another converging lens (f2= 40cm).

a. Calculate the final image distance relative to the last lens.

b. Calculate the magnification. Is the image upright or inverted? Is the image larger or smaller than the original image?

c. Carefully draw a scaled ray diagram of the Keplerian telescope. **** Your diagram should agree with your calculations from parts A and B. If it does not, go back and be more careful.

Explanation / Answer

Given,

o1 = 120 cm ; f1 = 40 cm ; D = 90 cm ; f2 = 40 cm

a)We know from lens equation that,

1/f = 1/i + 1/o

i = o x f / (o - f)

i1 = o1 x f1 / (o1 - f1) = 120 x 40 / (120 - 40) = 60 cm

o2 = 90 - 60 = 30 cm

i1 = o2 x f2 / ( o2 - f2) = 30 x 40 / (30 - 40) = -120 cm

Hence, final image is located at i2 = -120 cm

b)we know that magnification is given by: M = -i/o

m1 = -i1/o1 = -60/120 = -0.5

m2 = -i2/o2 = -(-120)/30 = 4

M = m1 x m2 =- 0.5 x 4 = -2

So the image formed is inverted and larger than the object.(by a factor of 2).

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