Animated Solution for Physics - Optics: If we need a magnification of 375 from a compound microscope of tube length 150 mm and an objective of focal length 5 mm, the focal length of the eyepiece should be close to

The microscopic world is a realm of hidden wonders, invisible to the naked eye. To unlock this universe, humanity invented the compound microscope—a brilliant piece of optical engineering that uses not one, but two lenses to achieve massive magnification. In this problem, we are tasked with designing a microscope that magnifies an object by a staggering 375 times. Let's dive into the physics of how this is achieved and unravel the mystery of the missing exact answer.

A standard compound microscope consists of two converging lenses. The first is the objective lens, placed very close to the specimen. It has a very short focal length ($f_o$) and creates a real, inverted, and magnified intermediate image inside the microscope tube.

The second lens is the eyepiece, which acts like a simple magnifying glass. You look through this lens. It takes the intermediate image created by the objective and magnifies it even further, creating a massive, virtual final image. The distance between these two lenses is related to the tube length ($L$).

When we use a microscope for extended periods, we want our eyes to be relaxed. This happens when the final image is formed at infinity, a state known as normal adjustment.

The total magnifying power ($m$) of the microscope is the product of the linear magnification of the objective ($m_o$) and the angular magnification of the eyepiece ($m_e$).

Through geometric optics, we derive the standard approximation formula for normal adjustment:

Here, $D$ is the least distance of distinct vision, which is the closest distance a normal human eye can focus comfortably. By standard convention, $D=25\text{ cm}$.

Let's gather the parameters provided in the problem. We must be extremely careful with units, ensuring everything is in millimeters to avoid silly mistakes.

- Total magnification, $m=375$ - Tube length, $L=150\text{ mm}$ - Objective focal length, $f_o=5\text{ mm}$ - Least distance of distinct vision, $D=25\text{ cm}=250\text{ mm}$

Our goal is to find the focal length of the eyepiece, $f_e$.

We substitute our known values into the master equation:

First, let's simplify the fraction on the right side. Dividing $150$ by $5$ gives us $30$:

Multiplying $30$ by $250$ yields $7500$:

Now, we isolate $f_e$ by swapping it with $375$:

Performing the division, we find:

We calculated exactly $20\text{ mm}$. However, looking at the options provided—(a) $22\text{ mm}$, (b) $2\text{ mm}$, (c) $12\text{ mm}$, (d) $33\text{ mm}$—our exact answer is missing!

Don't panic; this is a classic scenario in physics problems involving optical instruments. The formula $m=\frac{L\cdot D}{f_o\cdot f_e}$ is actually an approximation. The true formula for magnification is $m=\frac{v_o}{u_o}\cdot \frac{D}{f_e}$. We approximate the image distance $v_o$ as the tube length $L$, and the object distance $u_o$ as the focal length $f_o$.

Because of these inherent approximations in the standard textbook formula, the calculated value often deviates slightly from the rigorous real-world value. In such cases, we must choose the option that is closest to our calculated result.

Comparing $20\text{ mm}$ to the options, $22\text{ mm}$ is the clear winner. Thus, the focal length of the eyepiece should be close to $22\text{ mm}$.