In the figure below, the converging lens\' focal length is 12.2 cm, d = 12.1 cm,

Question

In the figure below, the converging lens' focal length is 12.2 cm, d = 12.1 cm, and the diverging lens' focal length is -10.8 cm. A 1.3 cm tall object is 6.3 cm to the left of the converging lens. Image:http://www.webassign.net/userimages/Two Lenses.jpg?db=v4net&id;=125802 (a) Where is the image formed by the converging lens, measured from that lens (using negative for a virtual image)? cm (b) How tall is the image formed by the converging lens (using negative for an inverted image)? cm (c) Where is the image formed by the diverging lens, measured from that lens (using negative for a virtual image)? cm (d) How tall is the image formed by the diverging lens (using negative for an inverted image)? cm

Explanation / Answer

Part A)

Apply the lens equation

1/f = 1/p + 1/q

1/12.2 = 1/6.3 + 1/q

q = -13.0 cm

Thus the image is virtual and formed 13.0 cm to the left of the lens.

Part B)

Apply h'/h = -q/p

h'/1.3 = -(-13)/6.3

h' = 2.69 cm

Part C)

Since the diverging lens uses the image from the converging lens, it is 13.0 + 12.1 = 25.1 cm away from it.

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

1/-10.8 = 1/25.1 + 1/q

q = -7.55 cm

Thus the image is virtual and 7.55 cm to the left of that lens.

Part D)

Apply h'/h = -q/p

h'/2.69 = -(-7.55)/25.1

h' = .809 cm

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