Question

Two converging lenses, each of focal length 14.9 cm, are placed 39.3 cm apart, and an object is placed 30.0 cm in front of the first lens. Where is the final image formed?

The image is located
cm  ---Location--- behind the second lens. in front of the first lens. in front of the second lens.

What is the magnification of the system?
M =  ?

Answer 1

The distance of the first image from the first lens is computed using the formula

1/s + 1/s' = 1/f

where s is the distance of the object = 30.0 cm,

s' is the distance of the image

and f is the focal length of the first lens = 14.9 cm

So 1/30 + 1/s' = 1/14.9

or 1/s' = 1/14.9 - 1/30

which gives s' = 29.57 cm, the distance of the first image from the first lens.

The magnification of the first lens = -s'/s = - 29.57 /30.0 = -0.985

The negative sign says that the image of the first lens is inverted.

The image of the first lens is the object of the second lens which is 39.3 cm away.

Therefore the object of the second lens is 39.3 - 29.57 cm = 9.73 cm in front of the second lens.

So, for the second lens, s2 = 9.73 cm

and the equation becomes:

1/9.73 + 1/s2' = 1/14.9 where s2' is the image of the second lens

So, 1/s2' = 1/14.9 - 1/9.73

or s2' = -28.04 cm

The negative s2' implies that the second and final image is formed in front of the second

lens and between the first and second lenses.

The magnification of the second lens = -s2'/s2 = -(-28.04)/10 = 2.804

The overall magnification of the system = M1 * M2 = -0.985 * 2.804 = -2.762

The negative sign says that the final image is inverted. M =2.762