Help with microscope/telescope problems

[eku_girl83]

This is part of a web assignment due tomorrow night, so I need help fairly quickly :)
Ok...I know the equations to use for these problems, I'm just really having trouble figuring out what's what from the wording of the problem.
1) The focal length of the eyepiece of a certain microscope is 16mm. The focal length of the objective is 6 mm. The distance between the objective and eyepiece is 19.7 cm. The final image formed by the eyepiece is at infinity. Treat all lenses as thin.
a) What is the distance from the objective to the object being viewed? Looking for s, but I have no idea how to do this...
1/f=1/s+1/s' but what are the focal length and the image distance?
I suppose s' would be negative, since image is virtual?
b) What is the magnitude of the linear magnification produced by the objective?
m=-s'/s
again, I need to find s' and s
c) What is the overal angular magnification of the microscope?
25s'/f1f2
Again, need s'

Ok here's the second one:
A telescope is constructed from two lenses with focal lengths of 96 cm and 12 cm, the 96 cm lens being used as the objective. Both the object being viewed and the final image are at infinity.
a) Find the angular magnification for the telescope.
I calculated this (correctly) to be -8.
b) Find the height of image formed by the objective of a building 60 m tall, 3 km away.
c) What is the angular size of the final image as viewed by an eye very close to the eyepiece?
No clue on parts b and c, so any insight would be appreciated...

Everyone on this forum has always been really helpful to me in the past. Thanks again!

 

[member 553083]

For part a, the angular magnification is given by M = (f1*f2)/(s'*s), where f1 and f2 are the focal lengths of the two lenses, and s and s' are the object distance and image distance respectively. In this case, both s and s' are assumed to be infinity, so the angular magnification reduces to M = f1/f2 = 8. For part b, the height of the image formed by the objective can be calculated using the equation h' = h*m, where h is the height of the object, m is the linear magnification of the objective, and h' is the height of the image. The linear magnification can be calculated using the equation m = -s'/s, where s' is the image distance and s is the object distance. In this case, s is 3km and s' is -96cm. Thus, m = -96/(-3000) = 0.032. Substituting this value into the equation gives h' = 60*0.032 = 1.92 m.For part c, the angular size of the final image viewed by an eye close to the eyepiece can be calculated using the equation θ = h'/d, where h' is the height of the image and d is the distance from the eyepiece to the eye. Since the eyepiece is assumed to be very close to the eye, we can assume that d = 0, giving θ = h'/0 = ∞.

 

[M98Ranger]

Hi! I'm sorry to hear that you're having trouble with these problems. I'll try my best to help you out.

For the first problem, it's important to remember that the focal length of a lens is the distance from the lens to its focal point. In this case, the objective has a focal length of 6 mm and the eyepiece has a focal length of 16 mm. The distance between the objective and eyepiece is given as 19.7 cm.

To find the distance from the objective to the object being viewed (represented by s in the equation), we can use the equation 1/f = 1/s + 1/s'. In this case, we know the focal length of the objective (6 mm) and the focal length of the eyepiece (16 mm). We also know that the final image is formed at infinity, which means that the image distance (s') is very large. Therefore, we can assume that 1/s' is equal to 0. Plugging these values into the equation, we get 1/6 = 1/s + 0. Solving for s, we get s = 6 mm.

For part b, we can use the equation m = -s'/s to find the magnitude of the linear magnification produced by the objective. We already know the value of s from part a (6 mm). To find s', we can use the equation s' = -fs/f + f, where f is the focal length of the eyepiece (16 mm) and s is the distance from the objective to the eyepiece (19.7 cm). Plugging in these values, we get s' = -(16 mm)(19.7 cm)/(6 mm) + 16 mm = -52.2 cm. Now we can plug these values into the equation for magnification to get m = -(-52.2 cm)/(6 mm) = 8.7.

For part c, the overall angular magnification of the microscope can be calculated using the equation 25s'/f1f2. We already know the value of s' (52.2 cm) and the focal lengths of the objective (6 mm) and eyepiece (16 mm). Plugging these values into the equation, we get 25(52.2 cm)/(6 mm)(16 mm) = 27.4.

For