At what magnification `T` of a telescope with a diameter of the objective `D = 6.0 cm` is the illuminance of the image of an object on the retina not less than without the telescope ? The pupil diameter is assumed do be equal to `d_(0) = 3.0 mm`. The losses of light in the telescope are negligible.

Explanation

If `L` is the luminance of the object, `A` is its area, `s =` distance of the object then light falling on the objective is
`(Lpi D^(2))/(4s^(2)) A`
The area of the image formed by the telescope (assuming that the image concides with the object) is `T^(2) A` and the area of the final image on the retina is
`= ((f)/(s))^(2) T^(2) A`
Where `f =` focal length of the eye lens. thus the illuminance of the image on the retain (when the object is observed thorugh the telescope) is
`(LpiD^(2)A)/(4u^(2) ((f)/(s))^(2)T^(2)A) = (Lpi D^(2))/(4f^(2)T^(2))`
When the object is viewed directly, the illuminance is, similarly, `(L pi d_(0)^(2))/(4f^(2))`
We want `(L pi D^(2))/(4f^(2)T^(2)) ge (Lpid_(0)^(2))/(4f^(2))`
So, `T le (D)/(d_(0)) = 20` on substituation of the values.