Question

w5a) Calculate the position of the final image in the Astronomical Telecope with the data provided. Estimate the position of the first image, realizing that the object is very far away. Then, use the Thin-Lens Formula to calculate the position of the final image from the second lens.

b) Compute the magnification of this telescope. as the image erect or inverted? What value indicates this orientation?

Objective:
300mm converging at 79.9cm

Eyepiece:
50mm converging at 44.2cm

Answer 1

Solution:

5a) The object is very far away (Infinity) . the Objective forms a real , inverted image , which lies just beyond its focal point .

This first image location becomes the object distance for the eye lens. The final image formed is always virtual , magnified and inverted with respect to the original object.

Object distance for the eye piece (2nd lens) is at the focal length of the eyepiece, since L = fo +fe , and fo is the image distance of the first image.

Using thin lens formula, 1/fe = 1/p +1/q, where f= fe = 44.2 cm; p = object distance = fe

then image distance q = [1/fe - 1/p ]^-1 =

5b) Final magnification , when the final image is formed at infinity is m = -fo/fe

= - 79.9 /44.2 = - 1.8

The negative sign indicates the final image is inverted with respect to the original object.