An object is placed 75.0 cm from a screen.
(a)Where should a converging lens of focal length 8.0 cm be placed to form an image on the screen?
shorter distance __cm from the screen
farther distance __cm from the screen
(b)Find the magnification of the lens.
magnification if placed at the shorter distance ___
magnification if placed at the farther distance ___
Let do = object distance
di = image distance
di + do = 75
1/di + 1/do = 1/f = 1/8
1/(75 - do) + 1/do = 1/8
Solve for do.
do + (75 - do) = (1/8)(do)(75-do)
75 = (1/8)do*(75-do)
600 = do*(75-do)
do^2 -75do + 600 = 0
do = (1/2)[75 +/-sqrt(5625-2400)]
= 37.5 +/-28.4
= 65.9 or 9.1 cm
di = 9.1 cm from lens when do = 65.9
di = 65.9 cm from lens when do = 9.1
(magnification = 7.24)
To determine the positions and magnifications of the image formed by a converging lens, we can use the lens formula and the magnification formula.
(a) To find the position where the lens should be placed to form an image on the screen, we need to use the lens formula:
1/f = 1/v - 1/u
where:
f is the focal length of the lens
v is the image distance from the lens
u is the object distance from the lens (given as 75.0 cm)
In this case, the focal length (f) is given as 8.0 cm. Plugging in these values into the lens formula, we get:
1/8.0 = 1/v - 1/75.0
Rearranging the equation, we can solve for v:
1/v = 1/8.0 + 1/75.0
1/v = (75 + 8)/600
1/v = 83/600
v = 600/83 cm
The image distance (v) is approximately 7.23 cm.
Therefore, the converging lens should be placed at a shorter distance of 7.23 cm from the screen.
To find the farther distance, we can simply subtract the image distance from the object distance:
Farther distance = Object distance - Image distance
Farther distance = 75.0 cm - 7.23 cm
Farther distance = 67.77 cm
Therefore, the converging lens should be placed at a farther distance of 67.77 cm from the screen.
(b) To find the magnification of the lens, we use the magnification formula:
Magnification (m) = -v/u
where:
v is the image distance from the lens (7.23 cm)
u is the object distance from the lens (75.0 cm)
Plugging in the values:
Magnification = -7.23 cm / 75.0 cm
Magnification = -0.0964
Therefore,
- The magnification of the lens if placed at the shorter distance is approximately -0.0964.
- The magnification of the lens if placed at the farther distance is also approximately -0.0964.
(a) To determine where the lens should be placed to form an image on the screen, we can use the lens formula:
1/f = 1/di - 1/do
where:
f = focal length of the lens
di = image distance
do = object distance
Given:
do = 75.0 cm
f = 8.0 cm
At the shorter distance, the image distance is equal to the object distance, so di = do.
Substituting the given values into the lens formula, we have:
1/8.0 = 1/do - 1/75.0
Simplifying the equation, we get:
1/do = 1/8.0 + 1/75.0
1/do = (75.0 + 8.0) / (8.0 * 75.0)
1/do = 83.0 / 600.0
1/do = 0.1383
Taking the reciprocal of both sides, we find:
do = 1 / 0.1383
do ≈ 7.23 cm
Therefore, the lens should be placed approximately 7.23 cm from the screen at the shorter distance.
To find the farther distance, we can use the lens formula again. Substituting the given values into the lens formula:
1/8.0 = 1/di - 1/75.0
Simplifying the equation, we get:
1/di = 1/8.0 + 1/75.0
1/di = (75.0 + 8.0) / (8.0 * 75.0)
1/di = 83.0 / 600.0
1/di = 0.1383
Taking the reciprocal of both sides, we find:
di = 1 / 0.1383
di ≈ 7.23 cm
Therefore, the lens should also be placed approximately 7.23 cm from the screen at the farther distance.
(b) The magnification of a lens can be calculated using the formula:
magnification = -di / do
Using the values we obtained from part (a), at the shorter distance:
magnification = -7.23 cm / 75.0 cm
magnification ≈ -0.096
Therefore, the magnification if the lens is placed at the shorter distance is approximately -0.096.
At the farther distance:
magnification = -7.23 cm / 75.0 cm
magnification ≈ -0.096
Therefore, the magnification if the lens is placed at the farther distance is also approximately -0.096.