Question

A 5.33-mm-high firefly sits on the axis of, and 11.1 cm in front of, the thin lens A, whose focal length is 5.13 cm. Behind lens A there is another thin lens, lens B, with focal length 28.1 cm. The two lenses share a common axis and are 57.9 cm apart. Is the image of the firefly that lens B forms real or virtual?How far from lens B is this image located (expressed as a positive number)?What is the height of this image (as a positive number)?Is this image upright or inverted with respect to the firefly?

Answer 1

Assuming both lenses are converging lenses,

Image formed by first lens will be behind the lens and real since object distance greater than focal length of lens 1.

1/v + 1/u =1/f

u = 11.1

f = 5.13

v = 9.54 cm behind the lens

since both lenses are seperated by distance of 57.9 cm,

distance of image formed by lens 1 from lens 2 = 57.9 - 9.54 = 48.36 cm

This will act as object for lens 2

since object distance is greater than focal length of lens, the image will form behind the mirror.

1/v +1/u = 1/f

u = 48.36

f = 28.1

v = 67.07 cm

The image is Real and located 67.07 cm behind the second lens

Magnification = - v/u for 1 lens

Since the final image is magnified twice

M = - v1/u1 * - v2/u2 = - 9.54/11.1 * - 67.7/48.36 = 1.19

The image formed is upright.

The height of firefly = 1.19 * 5.33= 6.35 mm