An astronomical telescope has an objective lens of 80cm and an eyepiece of 5cm focal length.If the final image is formed 25cm from the eye piece when a distant object is viewed by the eye close to the eyepiece,determine the magnifying power of the telescope.What is the best position of the eye.

To determine the magnifying power of the telescope, we can use the formula:

Magnifying Power = (Final Image Distance) / (Least Distance of Distinct Vision)

The final image distance is given as 25 cm.

The least distance of distinct vision is the minimum distance at which the human eye can focus clearly, which is typically about 25 cm.

Therefore, the magnifying power of the telescope is:

Magnifying Power = 25 cm / 25 cm = 1

A magnifying power of 1 means that the telescope does not provide any additional magnification compared to viewing with the naked eye. It simply allows for a closer look at the object.

As for the best position of the eye, it should be placed at the position where the final image is formed, which in this case is 25 cm from the eyepiece. This will ensure that the image is focused and clear.