Question

The magnification produced by an astronomical telescope for normal adjustment is $10$ and the length of the telescope is $1.1\text{m}.$ The magnification when the image is formed at least distance of distinct vision is -
[Use, $D=25\text{cm}$]

• A. $14$
• B. $6$
• C. $16$
• D. $18$

Answer 1

The correct option is A $14$

Given,

$M=10\text{Length of Telescope}=L=f_0+f_e=1.1\text{m}$

$\because$ In normal adjustment, the magnification is given by,

$M=\frac{f_0}{f_e}$

$10=\frac{f_0}{f_e}\Rightarrow f_0=10f_e$

$\because \text{Length of Telescope}L=f_0+f_e=1.1$

$\Rightarrow 10f_e+f_e=1.1$

$\Rightarrow f_e=0.1\text{m}=10\text{cm}$

And $f_0=10f_e=1\text{m}=100\text{cm}$

Now, when final image is formed at $D$

$\left|\begin{array}{c}M\end{array}\right|=f_0(\frac{1}{D}+\frac{1}{f_e})=100(\frac{1}{25}+\frac{1}{10})=14$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.