In the figure below, the converging lens\' focal length is 14.4 cm, d = 13.6 cm,

Question

In the figure below, the converging lens' focal length is 14.4 cm, d = 13.6 cm, and the diverging lens' focal length is -14.1 cm. A 1.4 cm tall object is 6.1 cm to the left of the converging lens.(a) Where is the image formed by the converging lens, measured from that lens (using negative for a virtual image)?(b) How tall is the image formed by the converging lens (using negative for an inverted image)?(c) Where is the image formed by the diverging lens, measured from that lens (using negative for a virtual image)?(d) How tall is the image formed by the diverging lens (using negative for an inverted image)?

Explanation / Answer

Part A)

Apply 1/f = 1/p + 1/q

1/14.4 = 1/6.1 + 1/q

q = -10.58 cm

The negative means the image is formed 10.58 cm to the left of the converging lens and it is virtual.

Part B)

Apply h'/h = -q/p

h'/1.4 = -(-10.58)/(6.1)

h' = 2.43 cm tall

Part C)

Since the diverging lens is 13.6 cm from the converging lens, and the image from the first lens is 10.58 cm to the left of the lens. The object distance from the diverging lens is 10.58 + 13.6 = 24.18 cm away.

Apply 1/f = 1/p + 1/q

1/-14.1 = 1/24.18 + 1/q

q = -8.91 cm

The negative means the image is formed 8.91 cm to the left of the diverging lens and it is virtual.

Part D)

Apply h'/h = -q/p

h'/2.43 = -(-8.91)/(24.18)

h' = .895 cm tall

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