An astronomical telescope has an object lens of 80cm and an eyepiece of 5cm focal length.If the image is formed 25cm from the eye piece when a distant object is viewed by the eye close to the eye piece ,detemine the magnifying power of the telescope.What is the best position of the eye?
I am human
I have some issues with that question, and am not going to answer it they way they want.
It is much comfortable to have the objective image at the focal point of the eyepiece, so that the eye sees an image at infinity. In that case, the magnification of the telescope is obtained from the formula
M = f(obj)/f(eyepiece)
which equals 16, in this case. That is not much of a telescope. A true astronomical telescope will have a longer objective focus longer than 80 cm and an eyepiece with a shorter focal length than 5 cm, unless you are just using binoculars.
If y=e(power x)+ e(power-x)/2. find x when y=1
Post that as a separate question, with subject math
To determine the magnifying power of the telescope, we can use the formula for magnification:
Magnification = (Image Distance)/(Object Distance)
First, let's find the object distance. In this case, the distant object is viewed by the eye close to the eyepiece, so the object distance is effectively at infinity.
Next, let's find the image distance. Given that the image is formed 25cm from the eyepiece, we need to find where the image is formed by the objective lens. To do this, we can use the lens formula:
(1/f_object) = (1/v_object) - (1/u_object)
where f_object is the focal length of the objective lens, v_object is the image distance from the objective lens, and u_object is the object distance from the objective lens.
Since the object distance is at infinity, we can simplify the equation as:
(1/f_object) = (1/v_object)
Rearranging the equation, we find:
v_object = f_object
Given that the focal length of the objective lens is 80 cm, the image distance from the objective lens is also 80 cm.
Now, let's calculate the magnification using the formula:
Magnification = (Image Distance)/(Object Distance)
= (25 cm)/(80 cm)
= 0.3125
The magnifying power of the telescope is approximately 0.31x.
As for the best position of the eye, it is recommended to position the eye at the point where the image formed by the eyepiece is at infinity. In this case, since the image is formed 25 cm from the eyepiece, the eye should be placed at a distance of 25 cm from the eyepiece to get the best view.