GOAL Calculate geometric quantities associated with a converging lens. PROBLEM A
ID: 1635768 • Letter: G
Question
GOAL Calculate geometric quantities associated with a converging lens.
PROBLEM A converging lens of focal length 10.0 cm forms images of an object situated at various distances. (a) If the object is placed 30.0 cm from the lens, locate the image, state whether it's real or virtual, and find its magnification. (b) Repeat the problem when the object is at 10.0 cm and (c) again when the object is 5.00 cm from the lens.
STRATEGY All three problems require only substitution into the thin-lens equation and the associated magnification equation. The conventions of signs for thin lenses must be followed.
SOLUTION
(A) Find the image distance and describe the image when the object is placed at 30.0 cm.
The ray diagram is shown in figure a. Substitute values into the thin-lens equation to locate the image.
Solve for q, the image distance. It's positive, so the image is real and on the far side of the lens:
q = +15.0 cm
The magnification of the lens is obtained from the relevant equation. Mis negative and less than 1 in absolute value, so the image is inverted and smaller than the object:
(B) Repeat the problem, when the object is placed at 10.0 cm.
Locate the image by substituting into the thin-lens equation:
This equation is satisfied only in the limit as q becomes infinite.
q
(C) Repeat the problem when the object is placed 5.00 cm from the lens.
See the ray diagram shown in figure b. Substitute into the thin-lens equation to locate the image.
Solve for q, which is negative, meaning the image is on the same side as the object and is virtual.
q = -10.0 cm
Substitute the values of p and q into the magnification equation. M is positive and larger than 1, so the image is upright and double the object size.
LEARN MORE
REMARKS The ability of a lens to magnify objects led to the inventions of reading glasses, microscopes, and telescopes.
QUESTION If the lens is used to form an image of the sun on a screen, the lens should be located at a distance from the screen equal to:
twice the distance from the lens to its focal point.the sum of the radii of the two lens surfaces. the radius of the surface of the lens.the distance from the lens to its focal point.a value between the distance to the focal point and twice the distance to the focal point.
PRACTICE IT
Use the worked example above to help you solve this problem. A converging lens of focal length 8.6 cm forms images of an object situated at various distances.(a) If the object is placed 25.8 cm from the lens, locate the image, state whether it's real or virtual, and find its magnification. (If either of the quantities evaluate to infinity, type INFINITY.)
(b) Repeat the problem when the object is at 8.6 cm.
(c) Repeat again when the object is 4.30 cm from the lens.
EXERCISE HINTS: GETTING STARTED | I'M STUCK!
Suppose the image of an object is upright and magnified 1.74 times when the object is placed 12.9 cm from a lens. Find the location of the image and the focal length of the lens.
1 + 1 = 1 p q fExplanation / Answer
A. given, focal length, f = 10 cm ( positive sign is for the converging lens)
a) object distance, u = -30 cm
let the image distance v = ?
then from thin lens formula
1/v - 1/u = 1/f
1/v = 1/10 - 1/30
v = 15 cm
as the image is behind the lens, it is real and inverted
magnification m = v/u = -15/30= -o.5
b) object distance, u = -10 cm
let the image distance v = ?
then from thin lens formula
1/v - 1/u = 1/f
1/v = 1/10 - 1/10
v = infinity
as the image is behind the lens, it is real and inverted
magnification m = v/u = -infinity
c) object distance, u = -5 cm
let the image distance v = ?
then from thin lens formula
1/v - 1/u = 1/f
1/v = 1/10 - 1/5
v = -10 cm
as the image is in front of the lens, it is virtual and erect
magnification m = v/u = 10/5 = 2
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