A converging lens of focal length 20.0 cm is separated by 50.0 cm from a converg

Question

A converging lens of focal length 20.0 cm is separated by 50.0 cm from a converging lens of focal length 5.00 cm. 1. Find the final position of the image of an object placed 40.0 cm in front of the first lens. 2. If the height of the object is 2.00 cm, what is the height of the final image? Is it real or virtual? 3. Determine the image position of an object placed 5.00 cm in front of the two lenses in contact. An object is located 36 cm to the left of biconvex lens of index of refraction 1.5. The left surface of the lens has a radius of curvature of 20 cm. The right surface of the lens is to be shaped so that a real image will be formed at 72 cm to the right of the lens. What is the required radius of curvature of the second surface?

Explanation / Answer

f1 = 20cm ; f2 = 5 cm ; d = 50 cm ;

1)from the lens eqn,

1/f = 1/i + 1/o

i = o x f / (o - f)

i1 = o1 x f1/(o1 - f1)

i1 = 40 x 20/(40 - 20) = 40 cm

now this image will serve as the object for second conversing lens placed. So,

o2 = 50 - 40 = 10 cm

i2 = o2 x f2/(o2 - f2)

i2 = 10 x 5/(10 - 5) = 4 cm

Hence, the final image will be at, i2 = 4 cm.

2)m1 = -i1/o1 = -40/20 = -2

m2 = -i2/o2 = -4/5 = -0.8

M = m1 x 2 = -2 x -0.8 = +1.6

h' = 1.6 x 2 = 3.2 cm

Hence, height of final image is = 3.2 cm

3)The effective focal length will be goven by

1/F = 1/f1 + 1/f2

1/F = 1/20 + 1/5 => F = 4 cm

i = o x f/(o - f)

i = 5 x 4/(5 -4) = 20 cm

Hence, i = 20 cm

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