Question

Draw the ray diagram to show the image formation by a refracting telescope and write the expression for angular magnification for the telescope in normal adjustment.

Hint:

For a refracting telescope in normal adjustment: \[ \boxed{ L=f_o+f_e } \] and \[ \boxed{ M=-\frac{f_o}{f_e} } \] A larger focal length of the objective and a smaller focal length of the eyepiece give a larger magnifying power.

Answer 1

Concept: A refracting telescope consists of two convex lenses:
• An objective lens of large focal length \(f_o\) and large aperture.
• An eyepiece lens of small focal length \(f_e\) and small aperture. The telescope is used to observe distant objects. Since the object is at a very large distance (practically at infinity), the rays coming from the object are nearly parallel. The objective forms a real, inverted and diminished image of the distant object at its principal focus. This image acts as the object for the eyepiece. In the condition of normal adjustment, the final image is formed at infinity so that the observer can view it comfortably without any strain on the eyes. Ray Diagram:

Step 1: Condition for normal adjustment.
For normal adjustment, the real image formed by the objective lies at the first focal plane of the eyepiece. Hence, the separation between the objective and eyepiece is \[ \boxed{ L=f_o+f_e } \] where
• \(f_o\) is the focal length of the objective,
• \(f_e\) is the focal length of the eyepiece.

Step 2: Expression for angular magnification.
The angular magnification or magnifying power of a telescope is defined as \[ M=\frac{\text{Angle subtended by the final image at the eye}} {\text{Angle subtended by the object at the unaided eye}}. \] For a refracting telescope in normal adjustment, \[ \boxed{ M=-\frac{f_o}{f_e} } \] The negative sign indicates that the final image is inverted with respect to the object. Hence, the magnitude of angular magnification is \[ \boxed{ |M|=\frac{f_o}{f_e} } \]