A converging lens of focal length 8000 cm is 195 cm to the l

A converging lens of focal length 8.000 cm is 19.5 cm to the left of a diverging lens of focal length -6.00 cm. A coin is placed 12.4 cm to the left of the converging lens. (a) Find the location of the coin\'s final image relative to the diverging lens. (Include the sign of each answer. Enter a negative value if the image is to the left of the diverging lens.) 6.1846 Correct: Your answer is correct. cm (b) Find the magnification of the coin\'s final image.

Solution

given

focal length of cnverging lens = 8cm

distance of it from diverging lense = 19.5cm

focal length of diverging lense =-6cm

object distance =12.4cm

The image of the first lens becomes the object of the second

So 1/s\'1 + 1/s1 = 1/f.1.....=>1/s\'1 = 1/f 1- 1/s1 = 1/8 - 1/12.4 = 0.04435

Therefore s\'1 = 22.54cm

This means the object for the second lens is 22.54 - 19.5 = 3.04 cm behind the diverging lens...so s = -3.04

(a virtual object)

Now for the second lens we have 1/s\'2 + 1/s2 = 1/f2

So 1/s\'2 = 1/f 2- 1/s2 = 1/(-6) - 1/(-3.04) = 0.4956

So s\'2 = 2.0177cm

So the image forms 2.0177 cm to the right of the diverging lens

Final magnification = m1*m2 = s\'1/s1*s\'2/s2 = 22.54/12.4*2.0177/3.04 = 1.206