Derive the expression for the magnifying power of an astronomical telescope for distinct vision. In an astronomical telescope the focal length of the objective lens is 75cm and that of the eyepiece is 5cm. Calculate the magnifying power of the telescope for distinct vision. (d =

Question

Derive the expression for the magnifying power of an astronomical telescope for distinct vision. In an astronomical telescope the focal length of the objective lens is 75cm and that of the eyepiece is 5cm. Calculate the magnifying power of the telescope for distinct vision. (d = 25 cm) (3+2=5)

**OR**

Derive the expression for the magnifying power of a compound microscope for distinct vision. A compound microscope uses an objective lens of focal length 4 cm and eyepiece lens of focal length 10 cm. An object is placed at 6 cm from the objective lens. Calculate the magnifying power of the compound microscope for distinct vision. (d = 25cm) (3+2=5)

Accepted Answer

> [!TLDR]
> Deriving each instrument's magnifying power as the ratio of the angle subtended by the final image to that subtended by the object: telescope $M=\frac{f_o}{f_e}\left(1+\frac{f_e}{D}ight)=18$; OR compound microscope $M=\frac{v_o}{|u_o|}\left(1+\frac{D}{f_e}ight)=7$.
>
> **Astronomical telescope, distinct vision (final image at the near point $D$) — derivation:**
>
> The objective (focal length $f_o$) collects parallel rays from the distant object and forms a real, inverted image $A'B'$ in its focal plane. The eyepiece (focal length $f_e$) is placed so that $A'B'$ lies just inside its focus; acting as a magnifier it forms the final virtual image at the near point $D$.
>
> The magnifying power is $M=\dfrac{\beta}{\alpha}$, the ratio of the angle $\beta$ subtended at the eye by the final image to the angle $\alpha$ subtended at the unaided eye by the object. Since $A'B'$ subtends $\alpha$ at the objective and $\beta$ at the eyepiece,
> $$\alpha\approx\frac{A'B'}{f_o},\qquad \beta\approx\frac{A'B'}{|u_e|}\ \Rightarrow\ M=\frac{\beta}{\alpha}=\frac{f_o}{|u_e|}$$
> For the eyepiece forming a virtual image at $D$, the lens equation gives $\dfrac{1}{|u_e|}=\dfrac{1}{f_e}+\dfrac{1}{D}$, so
> $$M=f_o\left(\frac{1}{f_e}+\frac{1}{D}ight)=\frac{f_o}{f_e}\left(1+\frac{f_e}{D}ight)$$
>
> **Numerical:** $f_o=75$ cm, $f_e=5$ cm, $D=25$ cm:
> $$M=\frac{75}{5}\left(1+\frac{5}{25}ight)=15	imes1.2=18$$
> i.e. magnifying power $18$ (final image inverted).
> …

Concept understanding — Astronomical Telescope

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