How do I calculate the diameter of the sun's image using two lenses in optics?

[T7]

Hi,

I have an optics Q here I'm not quite sure about:-

a.) An astronomers uses a single thin lens of focal length 0.5m to form a sharp image of the sun on a white screen. If the angular width of the sun is 0.5 degrees, what is the diameter of the sun's image.

b.) At night the astronomer uses the same lens to collect light from some stars, but not the screen is no longer used, and the light it permitted to pass through where the screen had been and through an eyepiece lens that acts as a magnifying glass to view the stars' images. The eyepiece lens has a focal length of 20mm, and the pair of lenses is an astronomical telescope. Assuming that the instrument is to be used with a relaxed eye, determine from a suitably labelled ray diagram the angular magnification effected by the telescope's optics. Is the final image erect or inverted?​

For the first part I would say that, at that distance, the rays from the sun striking the lens will be parallel to one another and will converge at some point from the lens equal in distance to the focal length of the lens, forming an image of height h. I presume I could then say that tan 0.5 = h / f, so the diameter is ftan 0.5.

For the second part, however, another lens is introduced. Am I to suppose that the image formed by the first is striking the second at a distance of 0m (?!)

I'm not entirely sure where to go from there to finish off the Q. Any help appreciated.

Cheers.

 

[Mattara]

I can give some hints:

1. The first lense is the second lense's focal point
2. "To be used by a relaxed eye" means that b (the distanse from the stars to the lins should be \infty (infinity)
3. The astronomical telescope consist of 2 positive lenses and is also known as a Kepler telescope, try and google that for the ray diagram

 

[andrevdh]

1. The image of the sun will for in the focal plane of the lens. Also the angle that it subtends (the image) is the same angle as the real sun - see the attachment. So you are correct in saying that \tan(0.5^o)=\frac{d}{f}.

 

[andrevdh]

The rays from the star falling on the objective will actually be parallel to each other if the drawing was to scale. Again the image forms at the focal plane of the objective. The statement "used with a relaxed eye" implies that the final rays emerging from the eyepiece need to be parallel giving an image at infinity. The first image formed by the objective therefore needs to be in the focal plane of the eyepiece to accomplish this.

 

[T7]

Ah, now I see.

Thanks folks.