Draw a labelled ray diagram of a refracting telescope when it forms image of a distant object at infinity. Derive expression for its magnifying power.
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To achieve large angular magnification $m = -f_o / f_e$, the objective lens must have a very large focal length ($f_o$), while the eyepiece must have a small focal length ($f_e$).
Concept: • An astronomical refracting telescope consists of two convex lenses: an objective lens of large focal length $f_o$ and large aperture, and an eyepiece of small focal length $f_e$ and small aperture.
• In normal adjustment, final image is formed at infinity. Step 1: Ray Diagram
Parallel rays from a distant object enter objective lens at an angle $\alpha$.
Objective lens forms a real, inverted, and diminished image $A'B'$ in its focal plane ($f_o$).
For final image to be at infinity, $A'B'$ must lie exactly at the principal focus $F_e$ of eyepiece lens.
Eyepiece refractor forms parallel rays emerging out at angle $\beta$ into observer's eye. Step 2: Derivation of Magnifying Power
Magnifying power $m$ of telescope is defined as ratio of angle $\beta$ subtended at eye by final image to angle $\alpha$ subtended by object at eye:
\[ m = \frac{\beta}{\alpha} \]
Since angles $\alpha$ and $\beta$ are very small:
\[ \alpha \approx \tan \alpha = \frac{A'B'}{O B'} = \frac{A'B'}{f_o} \]
\[ \beta \approx \tan \beta = \frac{A'B'}{E B'} = \frac{A'B'}{-f_e} \]
Here $O B' = +f_o$ is focal length of objective, and $E B' = -f_e$ is focal length of eyepiece (using Cartesian sign convention). Step 3: Final Expression
Substitute expressions for $\beta$ and $\alpha$:
\[ m = \frac{\frac{A'B'}{-f_e}}{\frac{A'B'}{f_o}} = -\frac{f_o}{f_e} \]
Length of telescope tube in normal adjustment is $L = f_o + f_e$. Step 4: Conclusion
The magnifying power of a refracting telescope in normal adjustment is $m = -\frac{f_o}{f_e}$. The negative sign indicates that final image is inverted with respect to object.
