A converging lens is placed 25.5 cm to the right of a diverging lens of focal le

Question

A converging lens is placed 25.5 cm to the right of a diverging lens of focal length 15.0 cm. A beam of parallel light enters the diverging lens from the left, and the beam is again parallel when it emerges from the converging lens.

A. Where is the image from the diverging located lens relative to the dicerging lens? Please include the distance from the lens and whether the image is on the left or right of the lens.

B. Where is the object for the converging lens located relative to the converging lens? Please include the distance from the lens and whether the image is on the left or right of the lens. (Hint: the object for the converging lens is the same as the image for the diverging lens)

C. What is the focal length of the converging lens?

Explanation / Answer

  The focal length of a converging lens is positive, but for a diverging lens, it is negative, therefore
f = -15 cm
We now use the object image relation for a thin lens (concave or diverging) which is defined by:

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

where s is the object distance, s' is the image distance and f is the focal length, therefore

1/25.5cm + 1/s' = 1/-15cm

1/s=-0.066-0.03

s=-9.52m

solve for s' and get approx. -9.52cm, which means the image formed is located 9.52cm to the left of the diverging lens if the object is located 25.5cm to the left of the diverging lens.

The lateral magnification for a thin lens is defined by

m=-s'/s

therefore

m = -(-9.52cm) / 25.5cm = +0.37

The magnification is positive, which means the image formed is erect (pointing upwards) if the object is pointing upwards.

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