Question

The focal lengths of the converging and diverging lenses are +15 and -20 cm, respectively. The distance between them is 50 cm and the object is placed 10 cm to the left of the converging lens. Determine the location of the final image with respect to the diverging lens. Is this image real or virtual? Find the total magnification; is the image inverse or upright?

Answer 1

Here ,

focal length of converging lens , f1 = 15 cm

focal length of diverging lens , f2 = -20 cm

for the first lens ,

object distance , do = 10 cm

Using lens formula

1/f1 = 1/di1 + do

1/15 = 1/di1 + 1/10

solving for di1

di1 = -30 cm

Now , for the second lens ,

object distance , do= 30 + 50

do = 80 cm

using lens formula

1/-20 = 1/di + 1/80

di = -16 cm

the image is located at 16 cm left of the diverging lens

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the image is virtual

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total magnification = (30/15) * (16/80)

total magnification = 0.4

the total magnification is 0.4

the image is upright