A system of two lenses forms an image of an arrow at x = x 3 = 69 cm. The first
ID: 1301852 • Letter: A
Question
A system of two lenses forms an image of an arrow at x = x3 = 69 cm. The first lens is a diverging lens located at x = 0 and has a focal length of magnitude f1 = 10.2 cm. The second lens is located at x = x2 = 29.7 cm and has an unknown focal length. The tip of the object arrow is located at (x,y) = (xo, yo) = (-42.1 cm, 18.1 cm).
1)
What is x1, the x co-ordinate of image of the arrow formed by the first lens?
cm
2)
What is y1, the y co-ordinate of the image of the tip of the arrow formed by the first lens?
cm
3)
What is f2, the focal length of the second lens. If the lens is a converging lens, f2 is positive. If the lens is a diverging lens, f2 is negative.
cm
4)
What is y3, the y co-ordinate of the image of the tip of the arrow formed by the two lens system?
cm
5)
The positions of the two lenses are now interchnaged (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)
Apply 1/f = 1/p + 1/q
1/-10.2 = 1/42.1 + 1/q
q = -8.21
Thus x1 = -8.21 cm
Number 2)
h'/h = -q/p
h'/18.1 = -(-8.21)/42.1
h' = 3.53 cm
Thus y1 = 3.53 cm
Number 3)
For the second lens, the image distance is 69 - 29.7 = 39.3 cm
The object distance is 8.21 + 29.7 = 37.91
1/f = 1/p + 1/q
1/f = 1/37.91 + 1/39.3
f = 19.3 cm (Positive so converging)
Number 4)
h'/h = -q/p
h'/3.53 = -39.3/37.91
h' = -3.74
Thus y3 = -3.74 cm
Part 5)
For the fist lens 1/19.3 = 1/42.1 + 1/q
q = 35.6 cm
Since the lenses are 29.7 cm apart, this image is behind the second lens (virtual and negative)
35.6 - 29.7 = 5.94
1/-10.2 = 1/-5.94 + 1/q
q = 14.2 cm
Since q is positive, the image is real and inverted
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