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 ___
To determine where the converging lens should be placed and the magnification of the lens, we can use the lens formula and magnification formula.
a) To find where the lens should be placed, we can use the lens formula:
1/f = 1/v - 1/u
where:
f = focal length of the lens (given as 8.0 cm)
v = image distance from the lens (distance from the lens to the screen)
u = object distance from the lens (distance from the object to the lens)
Let's substitute the given values into the lens formula:
1/8.0 cm = 1/v - 1/75.0 cm
To solve for v, we need to rearrange the equation:
1/v = 1/8.0 cm + 1/75.0 cm
Now, we can calculate the value of 1/v:
1/v = (75.0 cm + 8.0 cm) / (8.0 cm * 75.0 cm)
1/v = 83.0 cm / (8.0 cm * 75.0 cm)
1/v = 83.0 cm / 600.0 cm^2
v = 600.0 cm^2 / 83.0 cm
v ≈ 7.23 cm
Therefore, if the object is placed 75.0 cm from the screen, the converging lens of focal length 8.0 cm should be placed at approximately 7.23 cm from the screen.
For the shorter distance, the lens should be placed 7.23 cm from the screen.
For the farther distance, the lens should be placed 7.23 cm from the screen.
b) To find the magnification of the lens, we can use the magnification formula:
magnification (m) = -v/u
Let's calculate the magnification for both distances:
For the shorter distance:
magnification (m) = -v/u = -7.23 cm / 75.0 cm ≈ -0.096
For the farther distance:
magnification (m) = -v/u = -7.23 cm / 75.0 cm ≈ -0.096
Therefore, the magnification of the lens is approximately -0.096 for both the shorter and farther distances.