The magnifying power of a telescope with tube length 60 cm is 5. What is the focal length of its eye piece?

For an astronomical telescope adjusted for a final image at infinity, we first recall the two standard relations:

1. The magnifying power (angular magnification) is given by the formula

$$M \;=\;\frac{f_o}{f_e}$$

where $$f_o$$ is the focal length of the objective and $$f_e$$ is the focal length of the eye-piece.

2. The “tube length” of the telescope, that is, the distance between the objective and the eye-piece when the telescope is in normal adjustment, is simply the sum of their focal lengths:

$$L \;=\;f_o \;+\;f_e.$$

Now we substitute the numerical data given in the question. The magnifying power is

$$M \;=\;5,$$

and the tube length is

$$L \;=\;60 \text{ cm}.$$

From the magnification formula we can express the objective’s focal length in terms of the eye-piece’s focal length:

$$f_o \;=\;M\,f_e.$$

So, replacing $$f_o$$ by $$M f_e$$ in the tube-length relation, we have

$$L \;=\;f_o + f_e \;=\;M f_e + f_e.$$

Factorising $$f_e$$ on the right-hand side gives

$$L \;=\;(M + 1)\,f_e.$$

Now we substitute the known values $$L = 60 \text{ cm}$$ and $$M = 5$$:

$$60 \;=\;(5 + 1)\,f_e.$$

That simplifies to

$$60 \;=\;6\,f_e.$$

Dividing both sides by 6 to isolate $$f_e$$, we obtain

$$f_e \;=\;\frac{60}{6} \;=\;10 \text{ cm}.$$

Thus, the focal length of the eye-piece of the telescope is $$10 \text{ cm}$$.

Hence, the correct answer is Option D.