Optical Instruments Question: A system of two lenses forms an image of an arrow

Question

Optical Instruments Question:

A system of two lenses forms an image of an arrow at x times X3 times 69.1 cm. The first lens is a diverging lens located at x = 0 and has a focal length of magnitude f1 = 10.1 cm. The second lens is located at x times X2 times 29.7 cm and has an unknown focal length. The tip of the object arrow is located at (x,y) times (Xo, y0) times (-42.1 on, 18.1 cm). Known that: X-coordinate of image of the arrow formed by the first lens= -8.15 cm y-coordinate of the image of the tip of the arrow formed by the first lens= 3.50cm What is f2, the focal length of the second lence If the lens is a converging lens, f2 is positive. If the lens is a diverging lens, f2 is negative. What is yi, the y-coordinate of the image of the tip of the arrow formed by the two lens system? The positions of the two lenses are now interchanged (i.e., the second lens is moved to x = 0 and the diverging lens is moved to x = X2 = 29.7 cm). What is the nature of the final image in this new system? Real and Inverted Real and Upright Virtual and Inverted Virtual and Upright

Explanation / Answer

Number 1)

For the first lens, we will apply 1/f = 1/p + 1/q

1/-10.1 = 1/42.1 + 1/q

q = -8.15 cm

Then since the second lens is at 29.7, the distance from the frist image is 29.7 + 8.15 = 37.8 cm

This is the image for the second lens, so

1/f = 1/p + 1/q

The image distance is 69.1 - 29.7 = 39.4

1/f = 1/37.8 + 1/39.4

f = 19.3 cm

So the focal length of f2 = 19.3 cm. It is a converging lens

Number 2)

The magnification can be used for this part

M = -q/p times -q/p for the two lenses

M = -(-8.15/42.1) times -(39.4/37.8)

M = -.202

Initial y position is 18.1, so 18.1 times -.202 = -3.65 cm

Number 3)

For the first lens

1/19.3 = 1/42.1 + 1/q

q = 35.6 cm

For the second lens

1/-10.1 = 1/-5.9 + 1/q

q = 14.2 cm

Since the final q is positive, the image is real

For the magnification...

M = -(35.6/42.1) times -(14.2/-5.9)

M = -2.04

Since the overall magnification is negative the image is inverted

So, choice A - It is Real and Inverted

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