Example: A system of two lenses is setup, where f1 = f2 = 15 cm, and the two lenses are separated by 60 cm. The object is located 25 cm in front of the first lens. The first lens in convex (biconvex) and the second is concave (biconcave).

a) Describe the image formed by the first lens.b) Describe the image formed by the second lens.

1𝑠1

+1

𝑠1′=−

1𝑓 1

a)→1

𝑠1′=−

1𝑓 1−1𝑠1

→𝑠1′=37.5𝑐𝑚 Real Image

𝑚=− 𝑠′

𝑠¿−1.5 Inverted

b)

s1 = 25 cmf1 = -15 cm f2 = 15 cm

1𝑠2

+1

𝑠2′=−

1𝑓 2

→1

𝑠2′=−

1𝑓 2−1𝑠2

→𝑠2′=−9𝑐𝑚 Virtual Image

𝑚=− 𝑠′

𝑠¿0.4 Inverted

A lens is used to image an object onto a screen. If the right half of the lens is covered,

1. the left half of the image disappears.2. the right half of the image disappears.3. the entire image disappears.4. the image becomes blurred.5. the image becomes fainter.

f2

f1

io

The figure below shows three parallel horizontal light rays moving from left to right, passing through a rectangular block of plexiglass and focused to a point by a plexiglass lens. The focal point is marked by a large dark dot.

If the rectangular block is rotated counterclockwise, the rays will be displaced upward, but continue on parallel to each other and horizontal, as seen in the figure above with the lens removed. Note that the vertical position of the focus (the dot) is along the lower of the three rays. Now suppose that the lens is left in place when the rectangular block is rotated. What will happen to the position of the focus? (1) The focus will move UPWARD. (2) The focus will move DOWNWARD. (3) The focus will remain in the same position.

Note that for the previous derivation we have used a slightly different sign convention and therefore several of these expression have changed accordingly.

The boxed in quantities are the pieces that have been modified by the introduction of a minus sign.