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

The Secret to Minimum Strain

Imagine you are peering through a compound microscope, trying to observe a tiny specimen. If you stare for too long, your eyes might start to hurt. But did you know there is a specific optical configuration where your eyes are completely relaxed? This is known as the state of minimum strain, and understanding the physics behind it is the key to solving this classic JEE problem.

When the problem states that the strain on the eye is minimum, it is giving us a massive hint about the final image. Our eyes are most relaxed when they are looking at objects far away—at infinity. Therefore, for minimum strain, the compound microscope must form its final image at infinity.

How does a lens form an image at infinity? The object must be placed exactly at its focal point. In a compound microscope, the "object" for the eyepiece is actually the intermediate image formed by the objective lens.

So, the golden rule for minimum strain is: The intermediate image must form exactly at the focal point of the eyepiece.

Let's break down the geometry of the microscope tube. We are given: - Tube length, $L=10$ cm - Focal length of the objective, $f_o=1$ cm - Focal length of the eyepiece, $f_e=5$ cm

The tube length $L$ is the total distance between the objective lens and the eyepiece. Since the intermediate image is formed at the focal point of the eyepiece, the distance from the intermediate image to the eyepiece is simply $f_e=5$ cm.

This allows us to easily find the image distance for the objective lens, $v_o$:

Now we shift our focus entirely to the objective lens. We know its focal length ($f_o=1$ cm) and we just found where it forms the image ($v_o=5$ cm). We need to find where the original object was placed ($u_o$).

We apply the standard lens formula:

Substituting our known values:

Rearranging to solve for $u_o$:

The negative sign simply indicates that the object is placed in front of the lens, adhering to standard sign conventions. The physical distance of the object from the lens is the magnitude, $|u_o|=\frac{5}{4}$ cm.

The problem states this distance is equal to $\frac{n}{40}$ cm. Equating the two gives us our final answer:

$$n=50$$

By understanding the physical meaning of "minimum strain," a seemingly complex optical instrument problem breaks down into a simple application of the lens formula!