Question

A lighted candle is placed 38.0 cm in front of a diverging lens. The light passes through the diverging lens and on to a converging lens of focal length 14.0 cm that is 6.0 cm from the diverging lens. The final image is real, inverted, and 42.0 cm beyond the converging lens. Find the focal length of the diverging lens.

Answer 1

Using lens formula: V u LU V Here, starting from the final image formed by the converging lens v- 42 cm - 14 Cm 42 cm 14 cm -21 cm The image distance for the diverging lens is v--21 cm +6 cnm -15 cm u38.0 cm From lens formula -15 cm 38.0 cm -24.78 cm Therefore the focal length of the diverging lens is - 24.78 cm

Answer 2

We have to start from the final image formed by the converging lens to find the focal length of lens.
Image distance is 42cm and focal length is 14 cm take right positive direction
then by the lens formula we get 1/v-1/u = 1/f
for a converging lens v and f are positive
so 1/u = 1/v-1/f = 1/42 - 1/14 and so u = -21 cm
on the left side of the converging lens
this is the image for the diverging lens which is 5 cm from the converging lens
so v for converging lens is -21 + 6 = -17 cm [+6 indicating the right side ]
u is - 38 cm and so by the lens formula
1/v - 1/u = 1/f
-1/17 + 1/38 = 1/f
so f of diverging lens, f = - 30.76 cm