Aconverging lens of focal length 8.00 cm is 20.0 cm to the left of a diverging lens of focal length f-6.00 cm. a coin is placed 12.0 cm to the left of the converging lens. find the location and the magnification of the coin's final image.

final image is 2.4 cm to the right of diverging lens and final magnification is - 1.2

Explanation:

Consider the converging lens first :

f = focal length of the converging lens = 8.00 cm

p = distance of the coin from the lens = 12.0 cm

q = distance of the image of the coin

Using the lens equation

1/f = 1/p + 1/q

1/8 = 1/12 + 1/q

q = 24 cm

So the converging lens forms the image of the coin 24 cm to its right.

magnification is given as

m = -q/p

m = - 24/12

m = -2

So the converging lens forms the image of the coin twice its original size and inverted.

This image formed by the converging lens behaves as object for the diverging lens.

Consider the diverging lens and the image of the coin formed by converging lens

f = focal length of the diverging lens = - 6.00 cm

p = distance of the image of coin from the lens = 24 - 20 = 4 cm to its right

q = distance of the final image of the coin

Using the lens equation

1/f = 1/p + 1/q

1/(-6) = 1/4 + 1/q

q = - 2.4 cm

So the final image is 2.4 cm to the right of diverging lens

final magnification is given as

m' = m (-q/p)

m' = (-2) (- (- 2.4)/4)

m' = - 1.2

answer: pseudoscience