Question

The magnifying power of a refracting type of astronomical telescope is $m$. If focal length of eyepiece is doubled then the magnifying power will become

• A. $2m$
• B. $\sqrt{2}m$
• C. $\frac{m}{2}$
• D. $\frac{m}{4}$

Hint:

Magnifying power is inversely proportional to the focal length of the eyepiece ($m \propto \frac{1}{f_e}$). Therefore, scaling the eyepiece focal length by any factor $k$ automatically scales the total magnification by a factor of $\frac{1}{k}$.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given an astronomical telescope with an initial magnifying power $m$. We need to find how this magnifying power updates when the focal length of the eyepiece lens is exactly doubled.

Step 2: Key Formula or Approach:
The magnifying power (angular magnification) of a refracting astronomical telescope in normal adjustment is defined by the ratio of the focal length of the objective lens ($f_o$) to the focal length of the eyepiece lens ($f_e$):
$$m = \frac{f_o}{f_e}$$

Step 3: Detailed Explanation:
Let the initial setup parameters yield a magnification magnitude of:
$$m = \frac{f_o}{f_e}$$ According to the problem, the focal length of the objective lens $f_o$ remains completely unchanged, while the new focal length of the eyepiece becomes:
$$f_e' = 2f_e$$ Substitute this new eyepiece parameter into the magnification formula to compute the updated magnifying power $m'$:
$$m' = \frac{f_o}{f_e'} = \frac{f_o}{2f_e}$$ We can separate this fraction to express it in terms of the initial magnification $m$:
$$m' = \frac{1}{2}\left(\frac{f_o}{f_e}\right) = \frac{m}{2}$$ Thus, doubling the eyepiece focal length cuts the total magnifying capacity precisely in half.

Step 4: Final Answer:
The new magnifying power becomes $\frac{m}{2}$, corresponding to option (C).