Question

A 2.05-cm-tall object is placed 30.0 cm to the left of a converging lens with a focal length f1 = 20.5 cm . A diverging lens, with a focal length f2 = -42.5 cm , is placed 30.0 cm to the right of the first lens.

Q:How tall is the final image of the object?

Answer #1

**let**

**h = 2.05 cm
u1 = 30 cm
f1 = 20.5 cm
let v1 is the image distance for the first lens.**

**use, 1/u1 + 1/v1 = 1/f1**

**1/v1 = 1/f1 - 1/u1**

**1/v1 = 1/20.5 - 1/30**

**v1 = 64.7 cm**

**magnification, m1 = -v1/u1 = -64.7/30 =
-2.157**

**for second lens**

**object distance, u2 = -64.7 + 30 = -34.7 cm
f2 = -42.5 cm**

**let v2 is the image distance fro the second
lens**

**use, 1/u2 + 1/v2 = 1/f2**

**1/v2 = 1/f2 - 1/u2**

**1/v2 = 1/(-42.5) - 1/(-34.7)**

**v2 = 189 cm**

**magnification, m2 = -v2/u2 = -189/(-34.7) =
5.45**

**overall magnifcation, M = m1*m2 = -2.157*5.45 =
-11.7**

**image height = |M|*object height**

**= 11.7*2.05**

**= 24.0 cm
<<<<<<<<<-------------------Answer**

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