Question

The distance between the objective lens and the eye lens of an astronomical telescope when adjusted for parallel light is $100\text{cm}.$ The measured value of the magnification is $19.$ The focal length of the objective lens and the eye lens are-

• A. $85\text{cm}$ and $5\text{cm}$
• B. $190\text{cm}$ and $10\text{cm}$
• C. $95\text{cm}$ and $5\text{cm}$
• D. $50\text{cm}$ and $50\text{cm}$

Answer 1

The correct option is C $95\text{cm}$ and $5\text{cm}$

Given,

$\text{Magnification}{M}_{\infty }=19;L=100\text{cm}$

$M_{\infty }=\frac{f_0}{f_e}......\left(1\right)$

$\Rightarrow 19=\frac{f_0}{f_e}$

$f_0=19f_e$

The length of the tube i.e., distance between the objective and eye piece lens is,

$L=f_0+f_e$

$100=19f_e+f_e$

$100=20f_e$

$\therefore f_e=5\text{cm}$

Putting the value of $f_e$ in (1) we get,

$19=\frac{f_0}{5}$

$f_0=95\text{cm}$

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