Two converging lenses, each of focal length 14.9 cm, are placed 39.9 cm apart, a

Question

Two converging lenses, each of focal length 14.9 cm, are placed 39.9 cm apart, and an object is placed 30.0 cm in front of the first lens. Where is the final image formed? An object is placed 70.5 cm from a screen. where should a converging lens of focal length 5.0 cm be placed to from a clear image on the screen? (give your answer to at least one decimal place.) shorter distance cm from the screen farther distance cm from the screen Find the magnification of the lens. magnification if placed at the shorter distance magnification if placed at the farther distance

Explanation / Answer

1. for first lens,

f1 = 14.9 cm

do1 = 30cm

using lens equation,

1/f1 = 1/do1 + 1/di1

1/14.9 = 1/30 + 1/di1

di1 = 29.6 cm

for second lens,

do1 = 39.9 - 29.6 = 10.3 cm

1/14.9 = 1/10.3 + 1/di2

di2 = - 33.36 cm

image is 33.36 cm from seond lens. (inbetween lense)

M = (di1/do1 )(di2/do2) = - 3.2

-------------------

di + do = 70.5 cm

and using lens equation,

1/f = 1/di + 1/do

1/5 = 1/di + 1 / (70.5 - di)

1/5 = (70.5) / (di (70.5 - di))

70.5di - di^2 = 352.5

di^2 - 70.5di + 352.5= 0

di = 65 cm and 5.42 cm .......Ans

M = - (5.42) / (70.5 - 5.42) = - 0.417

and   - (65) / (70.5 - 65) = - 11.8

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