Galileo's first telescope used a convex objective lens with a focal length f = 1.65 m and a concave eyepiece. When this telescope is focused on an infinitely distant object, and produces an infinitely distant image, its angular magnification is +3.11. What is the focal length of the eyepiece?
How far apart are the two lenses?
Please help!? I have been stuck on this problem and I don't know where to begin.
To solve this problem, we can use the formula for angular magnification, M:
M = -f_obj / f_eye
Where:
M = angular magnification
f_obj = focal length of the objective lens
f_eye = focal length of the eyepiece
Given that the angular magnification is +3.11, we can substitute it into the formula:
3.11 = -f_obj / f_eye
Now, we can rearrange the equation to solve for the focal length of the eyepiece:
f_eye = -f_obj / 3.11
Given that the focal length of the objective lens, f_obj, is 1.65 m, we can substitute it into the equation:
f_eye = -(1.65 m) / 3.11
Now, we can calculate the focal length of the eyepiece:
f_eye ≈ -0.53 m
Since the eyepiece is concave, the focal length is negative.
To find the distance between the two lenses, we can use the lens maker's formula:
1 / f_obj = (n - 1) * (1 / R1 - 1 / R2)
Where:
f_obj = focal length of the objective lens
n = refractive index of the material the lenses are made of
R1 = radius of curvature of the objective lens
R2 = radius of curvature of the eyepiece
Since both lenses are made of glass, with a refractive index of approximately 1.5, we can substitute this value into the formula:
1 / 1.65 = (1.5 - 1) * (1 / R1 - 1 / R2)
Simplifying the equation further:
0.6061 = 0.5 * (1 / R1 - 1 / R2)
Now, we need some additional information about the radii of curvature for each lens. Once you provide that, we can proceed with the calculation.
To solve this problem, we can use the lens formula and the magnification formula. Let's break it down step by step:
1. First, let's define the given parameters:
- Focal length of the objective lens (f1) = 1.65 m
- Magnification (m) = +3.11
2. The lens formula relates the focal length of a lens (f), the object distance (d1), and the image distance (d2):
1/f = 1/d1 + 1/d2
In this case, the object distance is assumed to be at infinity. So, 1/d1 = 0 (since d1 approaches infinity).
3. The angular magnification (m) can be calculated using the formula:
m = -d2/d1
Since d1 approaches infinity, we can rewrite the equation as:
m = -d2/0
m = -∞
Here, the negative sign indicates that the image formed is inverted.
4. Since the angular magnification is given as +3.11, the solution must be +∞. This implies that the image formed should be at infinity.
5. Given that the telescope produces an infinitely distant image, the object distance (d2) can be considered as (-Focal length of the objective lens).
So, d2 = -f1 = -1.65 m
6. Now, we can substitute the values in the lens formula and solve for the unknown values.
1/f = 1/d1 + 1/d2
1/f - 0 = 1/d2
1/f = 1/(-1.65)
f = -1.65
Here, we obtained a negative value for the focal length. This indicates that the eyepiece is a diverging (concave) lens.
7. Finally, to find the focal length of the eyepiece, we can take the absolute value of the negative focal length calculated in step 6.
focal length of the eyepiece (f2) = |-1.65| = 1.65
So, the focal length of the eyepiece is 1.65 m.
8. To find the distance between the two lenses, we add the focal length of the objective lens and the focal length of the eyepiece.
Distance between the lenses = f1 + f2
Distance between the lenses = 1.65 + 1.65 = 3.3 m
Therefore, the focal length of the eyepiece is 1.65 m, and the distance between the two lenses is 3.3 m.