A 100cmhigh object is placed 435 cm to the left of a converg

A 1.00-cm-high object is placed 4.35 cm to the left of a converging lens of focal length 8.40 cm. A diverging lens of focal length 16.00 cm is 6.00 cm to the right of the converging lens. Find the position and height of the final image.

Solution

for the conferging lens :
di = (do * f)/(do - f)
= (4.35 * 8.4)/(4.35 - 8.4) = -9.02 cm ( at front of the converging lans)

for the diverging lens :
do = 9.02 + 6 = 15.02 cm
di = (do * f)/(do - f)
= (15.02 * (-16))/(15.02 -(-16))
= -7.74 cm ( at front of the diverging lens) --> it\'s the position of the final image)

the height of the final image:
h\'\'/h = l di/dol * ldi\'/do\'l
h\'\'/1 = (9.02/4.35) * (7.74/15.02)
h\'\' = 1.06 cm