A 6.45-mm-high firefly sits on the axis of, and 10.7 cm in front of, the thin le

Question

A 6.45-mm-high firefly sits on the axis of, and 10.7 cm in front of, the thin lens A, whose focal length is 6.49 cm. Behind lens A there is another thin lens, lens B, with focal length 23.3 cm. The two lenses share a common axis and are 63.1 cm apart.

a)Is the image of the firefly that lens B forms real or virtual?

b)How far from lens B is this image located (expressed as a positive number)?

c)What is the height of this image (as a positive number)?

d) Upright or inverted with respect to the firefly?

Explanation / Answer

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 = 10.7

f = 6.49

v = 16.495 ? 16.5cm behind the lens.

since both lenses are seperated by distance of 63.1 cm,

distance of image formed by lens 1 from lens 2 = 63.1 - 16.5 = 46.6 cm

This will act as object for lens 2

since object distance is greater than focal length of lens 2 ( 46.6 > 23.3 ), the image will form behind the mirror.

1/v +1/u = 1/f

u = 46.6

f = 23.3

v = 46.6 cm

The image is Real and located 46.6 cm behind the second lens

Magnification = - v/u for 1 lens.

Since the final image is magnified twice

M = - v1/u1 * - v2/u2 = - 16.5/10.7 * - 46.6/46.6 = 1.542

The image formed is upright.

The height of firefly = 1.542 * 6.45 = 9.95 mm

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