Question

A terrestrial telescope has an objective lens and eyepiece of focal length $200\text{cm}$ and $4\text{cm}$ respectively. Separation between the objective lens and the eyepiece is $220\text{cm}$. The power of the erecting lens is $n^2\text{D}$. The value of $n$(integer only)

Answer 1

Given,

$f_0=200\text{cm};f_e=4\text{cm};\text{and}L=220\text{cm}$

Where, $f_0=\text{Focal length of the objective}{f}_e=\text{Focal length of the eyepiece}$

For normal adjustment, the length of the telescope is,

$L=f_0+4f+f_e$

Where, $f$ is the focal length of the erecting lens.

$220=200+4f+4$

$\Rightarrow 4f=220-204=16$

$\therefore f=4\text{cm}$

Now, the power of the erecting lens is,

$P=\frac{1}{f}=\frac{100}{4}\text{m}=25\text{D}$

$\Rightarrow P=5^2\text{D}$

Given, $P=n^2\text{D}$

$\therefore n=5$

Key Note: $\left(i\right)$ A terrestrial telescope is used to see object at long distance on the surface of earth. Hence, image should be erect. $\left(ii\right)$ A terrestrial telescope has an additional erecting lens is used to make the final image erect.