Q118 / 180 · JIPMER 2016 · Physics
The magnification produced by a astronomical telescope for normal adjustment is 10 and the length of the telescope is 1.1 m . The magnification, when the image is formed atleast distance of distinct vision is
- A6
- B18
- C16
- Correct answerD14
Solution
Given, $m=10$, length of telescope $=1.1 \mathrm{~m}$
We know that,
Magnification, $m=\frac{t_0}{t_e}$
$\begin{aligned} & 10=\frac{t_0}{t_e} \\ & f_0=10 f_e \\ & \because \quad f_e+f_0=1.1 \mathrm{~m} \\ & \Rightarrow \quad t_e+10 t_e=1.1 \mathrm{~m} \\ & {\left[\because f_0=10 f_e\right]} \\ & \Rightarrow \quad f_e(1+10)=1.1 \mathrm{~m} \\ & \Rightarrow \quad f_e=0.1 \mathrm{~m} \text { or } 10 \mathrm{~cm} \\ & \end{aligned}$
Magnification least distance of distinct vision,
$\begin{aligned} M_b & =\frac{f_o}{f_e}\left(1+\frac{f_e}{D}\right) \\ & =10\left(1+\frac{10}{25}\right) \quad[\because D=25 \mathrm{~cm}] \\ & =10 \times \frac{35}{25} \\ & =14\end{aligned}$
Hence, the correct answer is option: 'D'.