Question

Magnification produced by astronomical telescope for normal adjustment is $10$ and length of telescope is $1.1\text{m}$. The magnification when the image is formed at least distance of distinct vision is $(D=25\text{cm})$ ?

• A. $10$
• B. $12$
• C. $14$
• D. $16$

Answer 1

The correct option is C $14$

Given,

Magnification for normal adjustment, $m=10$

Length of telescope, $L=1.1\text{m}=110\text{cm}$

We know,

$m=\frac{f_0}{f_e}=10\Rightarrow f_0=10f_e.....\left(i\right)$

$L=f_0+f_e=110.....\left(ii\right)$

Using equation (i) and (ii),

$f_0=100\text{cm}$ and $f_e=10\text{cm}$

Now, magnification when image is formed at least distance of distinct vision $(D=25\text{cm})$.

$m=\frac{f_0}{f_e}(1+\frac{f_e}{D})$

$\Rightarrow m=10(1+\frac{10}{25})=14$

Hence, $\left(C\right)$ is the correct answer.

Keynote: Magnification of telescope increases when an image is formed at least distance of distinct vision as compare to normal adjustment.