Question

An object 2.00 cm high is placed 30.8 cm to the left of a converging lens having a focal length of 25.8 cm. A diverging lens having a focal length of −20.0 cm is placed 110 cm to the right of the converging lens. (Use the correct sign conventions for the following answers.)

(a) Determine the final position and magnification of the final image. (Give the final position as the image distance from the second lens.)
final position :__________ cm (to the left of the first lens, between the two lenses, to the right of the second lens)
magnification _________


(b) Is the image upright or inverted?

(c) Repeat parts (a) and (b) for the case where the second lens is a converging lens having a focal length of +20.0 cm.

image position : ________cm (to the left of the first lens, between the two lenses, to the right of the second lens)  
magnification: _______

Is the image upright or inverted?

Answer 1

a) For the converging lens:

1/d_o1 + 1/d_i1 = 1/f₁

=> 1/30.8 + 1/d_i1 = 1/25.8

=> d_i1 = 25.8 * 30.8 / (30.8 - 25.8) = 158.9 cm

For the diverging lens:

d_o2 = (110 - 158.9) cm = -48.9 cm

So,

1/d_o2 + 1/d_i2 = 1/f₂

=> 1/(-48.9) + 1/d_i2 = 1/(-20)

=> d_i2 = 20 * 48.9 / (20 - 48.9) = -33.8 cm

So, the final image is virtual and 33.8 cm to the left of diverging lens. Hence, it is in between the two lenses.

magnification, m = (-d_i1/d_o1)(-d_i2/d_o2) = d_i1d_i2/d_o1d_o2 = 158.9 * (-33.8) / [30.8 * (-48.9)] = 3.58

b) Since the magnification is positive, image is upright.

c) For the second converging lens:

d_o2 = (110 - 158.9) cm = -48.9 cm

So,

1/d_o2 + 1/d_i2 = 1/f₂

=> 1/(-48.9) + 1/d_i2 = 1/20

=> d_i2 = 20 * 48.9 / (20 + 48.9) = 14.2 cm

So, the final image is real and 14.2 cm to the right of the second converging lens. Hence,the final image is to the right of the second lens.

magnification, m = (-d_i1/d_o1)(-d_i2/d_o2) = d_i1d_i2/d_o1d_o2 = 158.9 * 14.2 / [30.8 * (-48.9)] = -1.50

Since the magnification is negative, image is inverted.