Question

A lens system consists of two lenses 14.0 cm apart. The first lens is convex (converging) and has a focal length of 6.72 cm. An object is placed 3.29 cm in front of the first lens and produces an image 7.00 cm behind the second lens.
Q1:What is the focal length of the second lens?
A:5.215 cm

Q2:What is the magnification of the final image?

A2:

Please provide A2 and show all work, I already have the correct answer to Q1(5.215cm)

Answer 1

From fig, u1 = objecy distance of first lens = - 3.29 cm

f1 = focal length of first lens = + 6.72 cm           ( f = +ve because of convex lens)

V1 = image distance formed by first lens

By lens formula,    1/f₁ = 1/ V1 - 1/u1

1/V1 = 1/ f₁ + 1/ u1

V1 = - 6.433 cm        ( V1 comes out -ve ie image is virtual)

this virtual image V1 now become object for second lens.

so u2 = V1 + 14 cm = - ( 20.44 cm)

V2 = final image distance = 7 cm

f2 = focal length of second lens = ?

1/ f2 = 1/ V2 - 1/ u2

f2 = + 5.21 cm             ( here f2 comes out positive ie second lens is also convex)

( B ) magnification of final image will be:

m = m1 * m2

m = ( - V1/ -u1) * ( V2/ - u2)

m = ( -6.433/ - 3.29) * ( 7 / -20.44)

m = - 0.67

Here magnification comes out -ve ie image formed is real.