help me with these questions! A lens for a 35-mm camera has a focal length given
ID: 1570395 • Letter: H
Question
help me with these questions!
A lens for a 35-mm camera has a focal length given by f = 45mm a. How close to the film should the lens be placed to form a sharp image of an object that is 5.0 m away? b. What is the magnification of the image on the film? With unaided vision, a librarian can focus only on objects that lie at distances between 5.0 m and 0.50 m. a. Which type of lens (converging or diverging) is needed to correct his nearsightedness? b. Which type of lens will correct his farsightedness? c. Find the focal length of the lens needed for each part of the bifocal eyeglass lenses that will give the librarian normal visual acuity from 25 cm out to infinity. The 25cm is measured from his eye A friend tells you that when she takes off her eyeglasses and holds them 21 cm above a printed page the image of the print is erect but enlarged to 1.5 times its actual size. a. What is the focal length of your friend's glasses? b. Is the image real or virtual? Are the lenses in the glasses concave or convex? Your friend is 1.6 m tall. a. When she stands 3.0m from you, what is the height of her image formed on the retina of your eye? (Consider the eye to consist of a thin lens 1.7 cm from the retina.) b. What is the height of her image when she is 4.0 m from you? A 4k movie screen is about 10m times 10m Assume there are 4k times 4k pixels on the screen. If the smallest image size that can be resolved on your retina is 5um, how far back to you have to sit so you can no longer resolve individual pixels?Explanation / Answer
1)
1/f = 1/di + 1/do
1/di = 1/do - 1/f
1/di= 1/0.045 - 1/5
di = 45 mm .........(a)
m = -di / do = -(45 x 10^-3) / 5 m
m = -0.009
2)
a)
A diverging lens will produce an image of a distant object within the librarian’s far point.
(b)
A converging lens will produce an image beyond the librarian’s near point of an object that is within the near point.
c)
diverging (distant objects):
P = 1/f = 1/di + 1/do
P = 1/ (0.25 - 0.020)
P = -0.20 diapters
converging (near objects):
diverging (distant objects):
P = 1/f = 1/di + 1/do
P = 1/ (0.25 - 0.020) + 1 / (0.50 - 0.020)
P = +2.3 diapters
3)
First find the image distance:
m = -di / do
di = -mdo
di = -1.5 x 0.21 = -32 cm
Use this to find the focal length:
1/f = 1/di + 1/do
1/f = -1/32 + 1/21
f = 63 cm
b)
Since di < 0, the image appears on the same side of the lens as the object, which means it is a virtual image. Since
f > 0, the lenses are convex.
4)
hi = mho = -diho / do
hi = -(0.017 x 1.6) / 3
hi = -9.1 mm
b)
hi = -(0.017 x 1.6) / 4
hi = 6.8 mm
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.