Question: Draw a ray diagram for the formation of the image of an object by an astronomical telescope, in normal adjustment. Obtain the expression for its magnifying power.
The correct answer is –
Here’s a ray diagram for the formation of the image of an object by an astronomical telescope in normal adjustment:
In the diagram, AB represents the object being viewed through the telescope. The objective lens is represented by O1, and the eyepiece is represented by O2. The rays from the object are shown as parallel to each other and to the principal axis of the lens. The objective lens forms a real and inverted image of the object at its focal point. The eyepiece then acts as a magnifying glass, forming a virtual and enlarged image of the real image formed by the objective lens.
The magnifying power of an astronomical telescope in normal adjustment is given by the formula:
M = (angular magnification due to objective) x (angular magnification due to eyepiece)
where angular magnification due to objective is given by:
M_o = (f_o)/(f_e)
and angular magnification due to eyepiece is given by:
M_e = (25 cm)/(D)
where f_o is the focal length of the objective lens, f_e is the focal length of the eyepiece, and D is the distance between the objective lens and the eyepiece.
To obtain the expression for the magnifying power, we substitute the values of M_o and M_e into the formula for the magnifying power:
M = (angular magnification due to objective) x (angular magnification due to eyepiece) = M_o x M_e = (f_o)/(f_e) x (25 cm)/(D)
Therefore, the magnifying power of an astronomical telescope in normal adjustment is given by the formula:
M = (f_o)/(f_e) x (25 cm)/(D)
