Two systems are formed from a converging lens and a diverging lens. The point la

Question

Two systems are formed from a converging lens and a diverging lens. The point labeled Fconverging is the focal point of the converging lens. An object is placed inside the focal point of lens 1 at a distance of 7.60 cm to the left of lens 1. The focal lengths of the converging and diverging lenses are 15.00 and -20.0 cm, respectively. The distance between the lenses is 15.0 cm. Determine the final image distance for each system, measured with respect to lens 2. Diagram a shows Fconverging on the far left, then object, then #1, then #2. Diagram b shows object on far left, then #1, then Fconverging, then #2.

Explanation / Answer

The image of the first lens serves as the object of the second lens. Let's find where the image of the first lens appears.

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

s' is the image distance, what we are solving for

s is the object distance, 7.6 cm (object distance is always positive)

f is the focal length, 15.00 cm

so (1/f - 1/s)-1 = s'

(1/15 - 1/7.6)-1 = -15.4 cm

Since the image distance is negative, this means that it is a virtual image to the left of the first lens. It doesn't matter that this is virtual, we still use it as the object for the second lens. This is what you would expect for an object inside the focal length for a convex lens, think magnifying glass.

Don't forget to add the distance between the lenses.

15.4cm + 15cm = 30.4 cm

30.4 is our new s for the second lens

s'=(1/f - 1/s)-1

s'=(1/-20 - 1/30.4)-1 = -12.1 cm

So the virtual image is 12.1 cm to the left of the second lens.

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