PART 1: A lens of focal length +15.0 cm is 7 cm to the left of a second lens of
ID: 1499472 • Letter: P
Question
PART 1:
A lens of focal length +15.0 cm is 7 cm to the left of a second lens of focal length ?15.0 cm.
(a) Where is the final image of an object that is 27 cm to the left of the positive lens?
The image formed by the lens system is located _________ cm in front of the second lens.
(b) Is the image real or virtual?
(b) How must you be oriented to see the image?
A farsighted eye is corrected by placing a converging lens in front of the eye. The lens will create a virtual image that is located at the near point (the closest an object can be and still be in focus) of the viewer when the object is held at a comfortable distance (usually taken to be 25 cm). If a person has a near point of 76 cm, what power reading glasses should be prescribed to treat this hyperopia?
_______ D
The final image is real because di, B 0. The final image is real because dl, B > 0.Explanation / Answer
f1 = 15.0 cm
f2 = - 15.0 cm
(a)
do = 27 cm
We know,
1/f = 1/di + 1/do
1/15.0 = 1/27 + 1/di
di = 33.75 cm
Now, do = 33.75 - 7.0 = 26.75 cm
Distance of object, do = - 26.75 cm
Again,
1/f2 = 1/di + 1/do
-1/15 = -1/26.75 + 1/di
di = -34.2 cm
Image formed is located 34.2 cm infront of the second lens.
The final image is virtual , Because, di,B < 0
To see the correct image, Viewer must look right through both the lenses.
Part 2,
1/f = 1/di + 1/do
1/f = 1/25 - 1/76
1/f = 2.68
Power of the lens, P = 2.68 D
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.