1. A photographer uses a camera lens with focal length of 50.0 mm to photograph a tree that is 4.60 m tall. She wishes to fill her 24.0-mm slide with the tree.
(a) How far from the lens should the film be?
1/o + 1/24 = 1/50
600 + 25o = 24o
o = -600/13 or -46.15 mm
Therefore, the lens should be behind the film by 46.15 mm.
(b) What is the magnification of the image?
24/-46.15 = -0.52
Therefore, the magnification of the image is -0.52x.
This doesn't seem right. Am I doing something wrong?
2. The objective of a microscope has a focal length of 1.90 cm. An object is placed on the stage, 2.25 cm from the lens. The ocular has a focal length of 2.40 cm and is located 14.0 cm from the objective. What is the magnification of the image?
I am not really sure what to do with this one.
Are you a student doing online school?
No I am not. Is there a reason you are asking me this?
Are you sure you're not in an online school and that you are at a physical school?
1. For the first question, it seems there might be an error in the calculation. Let's go through the problem again to find where the mistake occurred.
(a) The lens equation for a thin lens is:
1/f = 1/o + 1/i
where:
f is the focal length of the lens,
o is the object distance (distance of the tree from the lens), and
i is the image distance (distance from the lens to the film).
In this case, f = 50.0 mm (or 0.05 m) and o = 4.60 m. We need to find i.
Using the lens equation, we can rearrange it to solve for i:
1/i = 1/f - 1/o
Substituting the given values, we have:
1/i = 1/0.05 - 1/4.60
Calculating that expression, we get:
1/i ≈ 19.35 - 0.22 ≈ 19.13
Taking the reciprocal of both sides, we find:
i ≈ 1/19.13 ≈ 0.052 m (or 52 mm)
So, the film should be approximately 52 mm from the lens (or behind the lens).
(b) The magnification (m) of the image can be calculated using the formula:
m = -i/o
Substituting the values from our previous calculations, we have:
m ≈ -0.052/4.6 ≈ -0.011
Therefore, the correct magnification of the image is approximately -0.011x, which means the image is smaller than the object and inverted.
2. For the second question, we will use a similar approach.
Given:
Focal length of the objective (f1) = 1.90 cm (or 0.019 m).
Object distance (o) = 2.25 cm (or 0.0225 m).
Focal length of the ocular (f2) = 2.40 cm (or 0.0240 m).
Distance between the objective and ocular (d) = 14.0 cm (or 0.140 m).
Using the lens equation, we can find the position of the image formed by the objective lens:
1/i1 = 1/f1 - 1/o
Substituting the given values, we have:
1/i1 = 1/0.019 - 1/0.0225
Simplifying that expression, we get:
1/i1 ≈ 52.63 - 44.44 ≈ 8.19
Taking the reciprocal, we find:
i1 ≈ 1/8.19 ≈ 0.122 m (or 12.2 cm)
Now, using the lens equation for the eyepiece, we can find the final image formed by the ocular lens:
1/i2 = 1/f2 - 1/i1-d
Substituting the given values and i1 from our previous calculation, we have:
1/i2 = 1/0.024 - 1/0.122-0.140
Simplifying that expression, we get:
1/i2 ≈ 41.67 - 5.84 ≈ 35.83
Taking the reciprocal, we find:
i2 ≈ 1/35.83 ≈ 0.028 m (or 2.8 cm)
The magnification (m) of the image can be calculated using the formula:
m = -i2/o
Substituting the values obtained above, we have:
m ≈ -0.028/0.0225 ≈ -1.24
Therefore, the magnification of the image is approximately -1.24x, meaning the image is magnified and inverted.
If you have any further questions or need additional clarification, feel free to ask!