In a microscope of tube length 1O cm two convex lenses are arranged with focal length of 2 cm and 5 cm. Total magnification obtained with this system for normal adjustment is $$(5)^{k}$$ The value of k is ___.

A compound microscope has tube length $$L = 10$$ cm, objective focal length $$f_o = 2$$ cm, and eyepiece focal length $$f_e = 5$$ cm. We wish to find the exponent $$k$$ such that the total magnification is $$(5)^k$$.

In normal adjustment (image at infinity), the total magnification is given by $$M = \frac{L \times D}{f_o \times f_e},$$ where $$D = 25$$ cm is the least distance of distinct vision.

Substituting the given values, $$M = \frac{10 \times 25}{2 \times 5} = \frac{250}{10} = 25 = 5^2.$$

Since $$5^k = 25 = 5^2,$$ it follows that $$k = 2$$. Therefore, the correct answer is Option A: $$k = 2$$.