Question

A converging lens has a focal length of 87.0 cm. Locate the images for the following object distances, if they exist. Find the magnification. (Enter 0 in the q and M fields if no image exists.)

(a)    87.0 cm

q	= ______ cm
M	= _________

Select all that apply to part (a).

a) real

b)virtual

c)upright

d)inverted

e)no image

(b)    24.9 cm

q	=______ cm
M	= _____

Select all that apply to part (b).

a) real

b)virtual

c)upright

d)inverted

e)no image

(c)    348 cm

q	= _____ cm
M	= _______

Select all that apply to part (c).

a) real

b)virtual

c)upright

d)inverted

e)no image

Answer 1

a) The object is at the focal point of the lens and therefore light from the object emerges from the lens as a parallel beam. There is therefore no image.

b) The lens equation is
1/u + 1/v = 1/f
where u is the object distance, v is the image distance and f is the focal length. The convention I use is that distances to real objects and images are positive while distances to virtual objects and images are negative. The focal length of a convex lens is positive and the focal length of a concave lens is negative.
1/24.9 + 1/v = 1/87
1/v = 1/87 - 1/24.9 = -34.88
v = -34.88 cm
The image is virtual and upright.
Magnification = v/u = 34.88/24.9 = 0.129

(c) 1/u + 1/v = 1/f
1/348 + 1/v = 1/87
1/v = 1/87 - 1/348 = 0.068
v = 116 cm
The image is real and inverted.
magnification = v/u = 116/348 = 0.33