Question

A) A small telescope has an objective lens of focal length $140\text{cm}$ and an eyepiece of focal length $5.0\text{cm}$. What is the separation between the objective lens and the eyepiece?

B) A small telescope has an objective lens of focal length $140\text{cm}$ andan eyepiece of focal length $5.0\text{cm}$. If this telescope is used to view a $100m$ tall tower $3\text{km}$ away, what is the height of the image of the tower formed by the objective lens?

C) A small telescope has an objective lens of focal length $140\text{cm}$ and an eyepiece of focal length $5.0\text{cm}$. What is the height of the final image of the tower if it is formed at $25\text{cm}$? (Height of the image of the tower formed by the objective lens is $\frac{140}{30}\text{cm})$

Answer 1

A) Formula used: Length of telescope = $f_o+f_e$
Given,
focal length of the objective lens, $f_o=140\text{cm}$
Focal length of the eyepiece, $f_e=5\text{cm}$
Separation between the objective lens and the eyepiece, $=f_o+f_e=(140+5)\text{cm}$
Final answer: $145\text{cm}$.

B) Step 1: Find the angle subtended by the tower at the telescope.
Formula used: $\alpha =\frac{h_1}{u}$
Height of the tower, $h_1=100m$
Distance of the tower (object) from the telescope, $u=3\text{km}=3000m$
The angle subtended by the tower at the telescope is given by,
$\alpha =\frac{h_1}{u}=\frac{100}{3000}=\frac{1}{30}\text{rad}$
Step 2: Find the height of the image formed by the objective.
Given,
focal length of the objective, $f_o=140\text{cm}$
Let $h_2$ be the height of the image formed by the objective.
The angle subtended by the image produced by the objective lens,
$\alpha =\frac{h_2}{f_o}=\frac{h_2}{140}$
But the angle subtended by the tower at the telescope and the angle subtended by the image produced by the objective lens are equal.
Therefore, $\frac{1}{30}=\frac{h_2}{140}$
$h_2=\frac{140}{30}\approx 4.7\text{cm}$
Final answer: $4.7\text{cm}$.

C) Step 1: Find the magnification of the eyepiece.
Formula used: $m_e=(1+\frac{D}{f_e})$
Given,
focal length of the eyepiece, $f_e=5\text{cm}$
Final image is formed at distance,
$D=25\text{cm}$
When the final image is formed at D, the angular magnification produced by the eyepiece,
$m_e=(1+\frac{D}{f_e})$
$m_e=(1+\frac{25}{5})$
$m_e=6$

Step 2: Find the height of the final image.
Formula used: $m_e=\frac{h_3}{h_2}$
Given,
height of the image of the tower formed by the objective lens, $h_2=\frac{140}{30}\text{cm}$
This image will act as virtual object for the eyepiece,
Let the height of the final image be $h_3$.
From magnification
$m_e=\frac{h_3}{h_2}$
$h_3=m_e\times h_2$
$h_3=6\times \frac{140}{30}$
$h_3=28\text{cm}$
Final answer : $28\text{cm}$.