Question

two things lenses of focal lengths -15cm and +15cm are placed 12cm apart. The negative lens is positioned to the left of the positive lens. A 4mm tall object is placed 5cm to the left of the negative lens .calculate a) distance between object and final image created by both lenses
b) height of the final image

Answer 1

Using the lens equation for first diverging lens

1/f = 1/u + 1/v

u = object distance = +5 cm

f = focal length = -15 cm

v = image distance = ?

1/v = -1/15 - 1/5

v = -15*5/(15 + 5) = -3.75 cm

Now this image will be -3.75 cm left from the diverging lens, Image's distacne from converging lens will be

u1 = 12 - (-3.75) = +15.75 cm = object distnace for converging lens

f1 = focal length of converging lens = +15 cm

v1 = image distance = ?

1/v1 = 1/15 - 1/(-15.75)

v1 = 15*15.75/(15 + 15.75) = +7.68 cm

v1 = +7.68 cm (+ve sign means image to the right of converging lens)

d = distance between object and final image = 5 cm + 12 cm + 7.68 = 24.68 cm

Part B.

Magnification is given by:

M = M1*M2 = (-v/u)*(-v1/u1)

M = (-(-3.75)/5)*(-7.68/15.75)

M = -0.366

We know that

M = hi/ho = -0.366

ho = height of object = 4 mm

hi = M*ho = -0.366*4 = -1.46 mm

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