Question

Draw a labelled ray diagram of a refracting telescope when it forms image of a distant object at infinity. Derive the expression for its magnifying power.

Hint:

For a refracting telescope in normal adjustment: \[ L=f_o+f_e \] and \[ M=-\frac{f_o}{f_e} \] A large focal length objective and a small focal length eyepiece give a high magnifying power.

Answer 1

Concept: A refracting telescope is an optical instrument used to observe distant objects such as stars, planets and other celestial bodies. It consists of two convex lenses:
• Objective lens of large focal length \(f_o\) and large aperture.
• Eyepiece lens of small focal length \(f_e\). The objective collects light from a distant object and forms a real, inverted and diminished image near its focal plane. The eyepiece acts as a simple microscope and magnifies this image. For normal adjustment, the final image is formed at infinity. This arrangement is preferred because the eye remains relaxed while observing.

Step 1: Labelled ray diagram of a refracting telescope in normal adjustment.

Step 2: Formation of image by the objective lens. The object is assumed to be situated at a very large distance from the telescope. Hence, the rays coming from the object are nearly parallel to the principal axis. The objective lens forms a real image at its focal plane. Let \[ f_o = \text{focal length of objective} \] and \[ f_e = \text{focal length of eyepiece} \] The image formed by the objective acts as the object for the eyepiece.

Step 3: Condition for normal adjustment. In normal adjustment, the intermediate image formed by the objective lies at the first focal plane of the eyepiece. Therefore, the eyepiece produces the final image at infinity. The distance between the objective and eyepiece is \[ L=f_o+f_e \] This arrangement allows the observer to view the image comfortably with minimum strain on the eye.

Step 4: Define magnifying power. The magnifying power of a telescope is defined as the ratio of the angle subtended by the final image at the eye to the angle subtended by the object at the unaided eye. Thus, \[ M=\frac{\beta}{\alpha} \] where
• \(\alpha\) is the angle subtended by the object at the objective.
• \(\beta\) is the angle subtended by the final image at the eyepiece.

Step 5: Determine the angle subtended by the object. Let the height of the intermediate image formed by the objective be \(h\). For small angles, \[ \tan\alpha \approx \alpha \] Hence, \[ \alpha=\frac{h}{f_o} \]

Step 6: Determine the angle subtended by the final image. The intermediate image is placed at the focal point of the eyepiece. For small angles, \[ \tan\beta \approx \beta \] Therefore, \[ \beta=\frac{h}{f_e} \]

Step 7: Calculate the magnifying power. Using \[ M=\frac{\beta}{\alpha} \] we obtain \[ M=\frac{\dfrac{h}{f_e}}{\dfrac{h}{f_o}} \] \[ M=\frac{f_o}{f_e} \] Since the final image is inverted with respect to the object, a negative sign is introduced. Thus, \[ \boxed{M=-\frac{f_o}{f_e}} \] The negative sign indicates inversion of the image. Result: The magnifying power of a refracting telescope in normal adjustment is \[ \boxed{M=-\frac{f_o}{f_e}} \] and its magnitude is \[ \boxed{|M|=\frac{f_o}{f_e}} \]