1. A converging lens with a focal length of 6.60 cmforms an image of a 3.80 mm -
ID: 1502145 • Letter: 1
Question
1. A converging lens with a focal length of 6.60 cmforms an image of a 3.80 mm -tall real object that is to the left of the lens. The image is 2.10 cm tall and erect.
--- Where are the object and image located? givin as s, s' _____________ cm
2.The focal length of a simple magnifier is 7.70 cm . Assume the magnifier to be a thin lens placed very close to the eye.
A. How far in front of the magnifier should an object be placed if the image is formed at the observer's near point, 25.0 cm in front of her eye? IN CM
B. If the object is h = 1.30 mm high, what is the height of its image formed by the magnifier? IN MM
3.The objective lens and the eyepiece of a microscope are 16.5 cm apart. The objective lens has a magnification of 59 × and the eyepiece has a magnification of 11 ×. Assume that the image of the objective lies very close to the focal point of the eyepiece.
A.Calculate the overall magnification of the microscope. M=
B.Calculate the focal length of the eyepiece. F1 = (give units)
C.Calculate the focal length of the objective. F2= (give units)
The largest refracting telescope in the world is at Yerkes Observatory in Wisconsin. The objective lens is 1.02 m in diameter and has a focal length of 19.4 m . Suppose you want to magnify Jupiter, which is 1.38×105 km in diameter, so that its image subtends an angle of 12 (about the same as the moon) when it is 6.28×108 km from earth.
What focal-length eyepiece do you need? Fe= (in m)
Explanation / Answer
Magnification
M=-S'/S = hi/ho
=>-S'/S = 2.1/0.38 = 5.526
S'=-5.526S
From Thin lens formula
1/f = 1/S + 1/S'
1/6.6 = -1/5.526S + 1/S=0.819/S
S = 5.4 cm
S'=-29.9 cm
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