A diverging lens located in the y-z plane at x = 0 forms an image of an arrow at

Question

A diverging lens located in the y-z plane at x = 0 forms an image of an arrow at x = x2 = -36.2 cm. The image of the tip of the arrow is located at y = y2 = 2.2 cm. The magnitude of the focal length of the diverging lens is 54.1 cm.

1)

What is x1, the x co-ordinate of the object arrow?.cm

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2)

What is y1, the y co-ordinate of the tip of the object arrow?cm

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3)

A converging lens of focal length fconverging = 28.91 cm is now inserted at x = x3 = -67.83 cm. In the absence of the diverging lens, at what x co-ordinate, x4, would the image of the arrow form?cm

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4)

To determine the image of the arrow from the combined converging + diverging lens system, we take the image from the converging lens (in the absence of the diverging lens) to be the object for the diverging lens. If this image is downstream of the diverging lens, the object for the diverging lens is virtual.

All this means is that the rays entering the diverging lens are converging towards a location downstream of the diverging lens. The final image can be calculated using this virtual object distance and the focal length of the diverging lens.

If this description seems unclear to you, you can check out the more detailed description of multiple lens systems given in the Optical Instruments unit.

What is x5, the x co-ordinate of the final image of the combined system?

Explanation / Answer

Part 1)

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

1/-54.1 = 1/p + 1/-36.2

p = 109.4 cm

Thus x1 = -109.4 cm

Part 2)

y2/y1 = -q/p

2.2/y1 = -(-36.2/109.4)

y1 = 6.65 cm

Part 3)

1/f = 1/p + 1/q

p = 109.4 - 67.83 = 41.57

1/28.91 = 1/41.57 + 1/q

q = 94.9

So x4 = 94.9 - 67.83 = 27.1 cm

Part 4)

1/f = 1/p + 1/q

1/-54.1 = 1/-27.1 + 1/q

q = 54.3

Thus x5 = 54.3 cm

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