An object is 19 m to the left of a lens of focal length 088

An object is 1.9 m to the left of a lens of focal length 0.88 m. A second lens of focal length -3.2 m is 0.7 m to the right of the first lens. Find the distance between the object and the final image formed by the second lens. What is the overall magnification (with sign)? Is the final image real or virtual?

Solution

for first lens:

f1 = 0.88 m

object distance, do1 = 1.9m

Applying lens formula,

1/f = 1/di + 1/do

1/0.88 = 1/di1 + 1/1.9

di1 = 1.64 m

now for second lens:

f2 = - 3.2 m

object distance, do2 = 0.7 - 1.64 = - 0.939 m

1/(-3.2) = 1/(- 0.939) + 1/di2

di2 = 1.33 m

distance between di2 and do1 = 1.33 + 1.9 + 0.7 = 3.93 m .....Ans

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m = m1 m2

m1 = - (di1 / do1)   and m2 = - (di2 / do2)

m = (di1 / do1 ) (di2 / do2)

= (1.64 / 1.9) ( 1.33 / -0.939)

      = - 1.22 .....Ans

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image will be virtual.

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m is negative hence image will be inverted.

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