Question

An object 2.00 cm high is placed 45.3 cm to the left of a converging lens having a focal length of 40.3 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

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

magnification

image orientation

upright inverted

Answer 1

cos If the second lens is convegna lers then Be the to de boda de tot ass 118 92= 18.546 m so the image is towards the right of He lens and over all magnification is lens M=( 21 )= -0.59 since magnification is negative the image is invested. height of object his dem Position of the object P = 45.3cm focal length R = 40.3 cm P+ akt 40.3 45.3 V = 365. 118 cm distane ble two lenses Da 110 cm so the image is formed at a distance - 255.118 cm from the second lens which acts as object for the second focal length of secondlers fz = - 20cm Peter on the to t255.118 Pe = -21.7cm so the image is towards the sight of the second lens (b) magnification m=(20) 6 )= +0.63 since magnification is position in the the image is Domented up right.