The focal lengths of the objective and the eyepiece of a compound microscope are 2.0 cm and 3.0 cm r

Q: Why is the object distance for the eyepiece equal to its focal length?

The problem states that the final image is formed at infinity. For a converging lens (like the eyepiece) to form an image at infinity, the object must be placed exactly at its principal focus. Here, the 'object' for the eyepiece is the intermediate image formed by the objective lens.

Q: What exactly is the tube length of a microscope?

The tube length is the physical distance between the objective lens and the eyepiece lens. In this problem, it is given as 15.0 cm.

Q: Why do we subtract $u_e$ from $L$ to get $v_o$?

The intermediate image is formed between the two lenses. The total distance $L$ is the sum of the distance from the objective to the image ($v_o$) and the distance from the image to the eyepiece ($u_e$). Therefore, $v_o + u_e = L$, which means $v_o = L - u_e$.

Q: Does the negative sign of $u_o$ matter?

The negative sign is a result of the Cartesian sign convention, indicating that the object is placed in front of the lens (against the direction of incident light). When asked for the 'distance', we usually state the magnitude, which is 2.4 cm.

Q: What if the final image was formed at the near point instead of infinity?

If the final image was formed at the near point (usually $D = 25$ cm), you would use the lens formula for the eyepiece with $v_e = -25$ cm to find $u_e$. Then, you would proceed with $v_o = L - u_e$ as usual.

Question

The focal lengths of the objective and the eyepiece of a compound microscope are 2.0 cm and 3.0 cm respectively. The distance between the objective and the eyepiece is 15.0 cm. The final image formed by the eyepiece is at infinity. The two lenses are thin. The distance in cm of the object and the image produced by the objective, measured from the objective lens, are respectively

Options:

(A) 2.4 and 12.0

(B) 2.4 and 15.0

(C) 2.0 and 12.0

(D) 2.0 and 3.0

Sigma Percentile · Accepted Answer

### The Anatomy of a Compound Microscope

Imagine you are looking through a compound microscope. The entire setup relies on the beautiful coordination of two lenses: the **objective lens** (which faces the object) and the **eyepiece** (which you look through).

In this problem, we are given the focal lengths of both lenses: $f_o = 2.0$ cm and $f_e = 3.0$ cm. The total distance between them, often called the tube length, is $L = 15.0$ cm. The key to unlocking this problem lies in the phrase: *"The final image formed by the eyepiece is at infinity."*

### Decoding the "Image at Infinity" Condition

What does it mean physically when the final image is at infinity? For any convex lens to form an image at infinity, the object must be placed exactly at its focal point.

In a compound microscope, the objective lens forms an intermediate image ($I_1$), which then acts as the object for the eyepiece. Therefore, for the eyepiece to project the final image to infinity, this intermediate image must be located exactly at the focal point of the eyepiece.

Mathematically, this means the object distance for the eyepiece is equal to its focal length:
$$u_e = f_e = 3.0 	ext{ cm}$$

### Finding the Image Distance for the Objective

Now, let's look at the geometry of the setup. The total distance between the two lenses is $15.0$ cm. The intermediate image is formed between them, at a distance of $3.0$ cm from the eyepiece.

So, how far is this intermediate image from the objective lens? It's simply the total distance minus the distance to the eyepiece:
$$v_o = L - u_e$$
$$v_o = 15.0 - 3.0 = 12.0 	ext{ cm}$$

This $12.0$ cm is the image distance ($v_o$) for the objective lens.

### Applying the Lens Formula

We now have the focal length ($f_o = 2.0$ cm) and the image distance ($v_o = 12.0$ cm) for the objective lens. To find where the original object was placed ($u_o$), we invoke the standard thin lens formula:
$$\frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o}$$

Let's carefully substitute our known values into the equation:
$$\frac{1}{12.0} - \frac{1}{u_o} = \frac{1}{2.0}$$

Rearranging the terms to solve for $\frac{1}{u_o}$:
$$\frac{1}{u_o} = \frac{1}{12.0} - \frac{1}{2.0}$$

To subtract these fractions, we find a common denominator, which is $12.0$:
$$\frac{1}{u_o} = \frac{1 - 6}{12.0} = -\frac{5}{12.0}$$

Inverting both sides gives us the object distance:
$$u_o = -\frac{12.0}{5} = -2.4 	ext{ cm}$$

The negative sign simply confirms our sign convention: the object is placed $2.4$ cm in front of the objective lens.

### The Final Verdict

We have successfully determined both required distances. The distance of the object from the objective lens is **$2.4$ cm**, and the distance of the intermediate image produced by the objective is **$12.0$ cm**. This perfectly aligns with option (a).

Select Answer:

Visualized Solution

\text{Compound Microscope Setup}

\text{Condition for Final Image at Infinity}

\text{Image Distance for Objective}

\text{Lens Formula for Objective}

\text{Substituting Values}

\text{Calculating Object Distance}

\text{Final Conclusion}

The Sigma Insight: Optical Instruments

The Anatomy of a Compound Microscope

Decoding the "Image at Infinity" Condition

Finding the Image Distance for the Objective

Applying the Lens Formula

The Final Verdict

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