A telescope that has a 100 cm focal length objective and a 3.00 cm focal length eyepiece. Find the distance between the objective and eyepiece lenses in the telescope needed to produce a final image very far from the observer, where vision is most relaxed. Note that a telescope is normally used to view very distant objects.
I know how to figure the magnification, which is -100/3 = -33.33, but I'm not sure what to do from here... Thanks for your help!
the image is at infinity so d=fo+fe=100+3=103cm
Its simple dawg you just need to believe in yourself
To find the distance between the objective and eyepiece lenses in the telescope, we can use the formula for the focal length of a compound lens system.
The formula for the focal length of a compound lens system is:
1/f = 1/f1 + 1/f2 - d/f1f2
Where:
f is the focal length of the compound lens system
f1 is the focal length of the objective lens
f2 is the focal length of the eyepiece lens
d is the distance between the two lenses
From the given information, we know that f1 = 100 cm and f2 = 3.00 cm. We also want to find the distance d.
Substituting the known values into the formula, we get:
1/f = 1/100 + 1/3 - d/(100*3)
Since the final image is very far from the observer, the observer's eye is at the location of the final image. In this case, the image formed by the telescope is at infinity. The formula for the compound lens system then simplifies to:
1/f = 1/f1 + 1/f2
Substituting the known values and simplifying, we get:
1/f = 1/100 + 1/3
1/f = (3 + 100)/(100*3)
1/f = 103/300
f = 300/103 ≈ 2.9126 cm
Therefore, the distance between the objective and eyepiece lenses in the telescope needed to produce a final image at infinity is approximately 2.9126 cm.