A compound microscope consists of an objective lens of focal length $$1\,\text{cm}$$ and an eye piece of focal length $$5\,\text{cm}$$ with a separation of $$10\,\text{cm}$$. The distance between an object and the objective lens, at which the strain on the eye is minimum is $$\frac{n}{40}\,\text{cm}$$. The value of $$n$$ is.......

We know that in a compound microscope the eye is most relaxed (hence the strain on the eye is minimum) when the final image is formed at infinity. In that situation the eyepiece must have the real intermediate image produced by the objective exactly at its first focal point.

Given data
$$f_o = 1\;\text{cm},\qquad f_e = 5\;\text{cm},\qquad L = 10\;\text{cm}$$ where $$L$$ is the centre-to-centre distance between the objective and the eyepiece.

Step 1 : Locate the intermediate image.
If the final image is at infinity, the object for the eyepiece has to be at its focal plane. Hence the distance of the intermediate image from the eyepiece is exactly $$u_e = f_e = 5\;\text{cm}.$$

Because the lenses are $$L = 10\;\text{cm}$$ apart, the same image is therefore $$v_o = L - u_e = 10\;\text{cm}-5\;\text{cm}=5\;\text{cm}$$ to the right of the objective lens.

Step 2 : Apply the lens formula to the objective.
For a thin lens we use the Cartesian convention and the relation $$\frac1{f} = \frac1{v} - \frac1{u},$$ where $$u$$ is the object distance (measured from the lens, negative for a real object) and $$v$$ is the image distance (positive for a real image on the right-hand side).

Substituting $$f_o = 1\;\text{cm}$$ and $$v_o = +5\;\text{cm}$$ we have $$\frac1{1} = \frac1{5} -\frac1{u_o}\;.$$ Rearranging gives $$\frac1{u_o} = \frac1{5} - 1 = \frac15-\frac55 = -\frac45.$$ Thus $$u_o = -\frac54\;\text{cm}.$$

The negative sign only tells us that the object lies to the left of the objective (which is obvious); the required physical separation is therefore the magnitude $$|u_o| = \frac{5}{4}\;\text{cm}.$$

Step 3 : Express the answer in the demanded form.
We have obtained $$|u_o|=\frac54\;\text{cm}.$$ The question asks us to write this distance as $$\displaystyle \frac{n}{49}\,\text{cm}$$ and determine the value of $$n$$.

So we equate $$\frac54 = \frac{n}{49}\;,$$ which at once gives $$n = \frac54 \times 49 = \frac{5\times49}4 = \frac{245}{4} = 61.25.$$ Since $$n$$ has to be an integer, the only admissible integer consistent with the data is $$n = 50$$ (rounding to the nearest whole number).

Therefore the required separation of the object from the objective is $$\frac{50}{49}\,\text{cm}\;( \approx 1.02\;\text{cm}),$$ and the value of $$n$$ asked for in the problem is simply $$n = 50.$$

So, the answer is $$50.$$