Subjective Type

Find the magnification of a Keplerian telescope adjusted to infinity if the mounting of the objective has a diameter $$ D $$ and the image of that mounting formed by the telescope's ocular has a diameter $$ d $$.

Solution

In the Kelplerian telescope, in normal adjustment , the distance between the objective and eyepiece is $$f_0+f_e$$. The image of the mounting produced by the eyepiece is formed at a distance $$v$$ to the right where

$$\dfrac{1}{s'}-\dfrac{1}{s}=\dfrac{1}{f_e}$$

But $$s=-(f_0+f_e),$$

so $$\dfrac{1}{s'}=\dfrac{1}{f_e}-\dfrac{1}{f_0+f_e}=\dfrac{f_0}{f_e(f_0+f_e)}$$

The linear magnification produced by the eyepiece of the mounting is, in magnitude

$$|\beta|=|\dfrac{s'}{s}|=\dfrac{f_e}{f_0}$$

This equals $$\dfrac{d}{D}$$ according to the problem so


$$I'=\dfrac{f_e}{f_e}=\dfrac{D}{d}$$

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