An object viewed from a near point distance of 25 cm, using a microscopic lens with magnification 6,

Q: Why do we use the formula $M = 1 + D/f$ for the first case?

The problem states the object is viewed from a 'near point distance of 25 cm'. This implies the virtual image formed by the simple microscope is at the least distance of distinct vision ($D = 25$ cm). The standard formula for magnification in this specific case is $M = 1 + D/f$.

Q: Why is the tube length $L$ used directly in the numerator for the compound microscope?

The exact formula for a compound microscope is $M = (v_0/u_0) \times (D/f_e)$. However, in standard problems where the tube length $L$ is much larger than the focal lengths, we approximate the image distance of the objective ($v_0$) to be roughly equal to the tube length $L$. Hence, the formula simplifies to $M \approx \frac{L \cdot D}{f_0 \cdot f_e}$.

Q: Why is the magnification 12 in the second case?

The problem explicitly states that the new setup provides a 'total magnification double the earlier'. Since the earlier magnification was 6, the new magnification is $2 \times 6 = 12$.

Q: What does 'unresolved image' mean in the context of this question?

It's a practical context clue. It implies that the simple microscope (magnification of 6) isn't powerful enough to see the fine details of the object clearly, which motivates the transition to a compound microscope for higher magnification.

Q: Why do we convert 0.6 m to 60 cm?

Consistency of units is crucial in physics. Since the least distance of distinct vision ($D$) is given as 25 cm, and we calculated the objective focal length ($f_0$) in cm, we must convert the tube length from meters to centimeters ($0.6 \text{ m} = 60 \text{ cm}$) before substituting it into the formula.

Question

An object viewed from a near point distance of 25 cm, using a microscopic lens with magnification 6, gives an unresolved image. A resolved image is observed at infinite distance with a total magnification double the earlier using an eyepiece along with the given lens and a tube of length 0.6 m, if the focal length of the eyepiece is equal to ......... cm.

Sigma Percentile · Accepted Answer

This problem is a beautiful journey through the evolution of optical instruments. We start with a simple microscope and then upgrade it to a compound microscope to achieve higher magnification. Let's break down the physics and the math step-by-step.

### The Simple Microscope Setup

Imagine you are using a single convex lens as a simple microscope. The problem states that the image is formed at the near point, which is the least distance of distinct vision, denoted by $D$. For a normal human eye, $D = 25 	ext{ cm}$.

When a simple microscope forms an image at the near point, its magnifying power $M$ is given by the formula:

$$M = 1 + \frac{D}{f_0}$$

Here, $f_0$ is the focal length of the microscopic lens. We are given that the magnification $M = 6$. Let's substitute the known values into our equation:

$$6 = 1 + \frac{25}{f_0}$$

Subtracting 1 from both sides, we get:

$$5 = \frac{25}{f_0}$$

Solving for $f_0$, we find the focal length of our initial lens:

$$f_0 = 5 	ext{ cm}$$

### Transitioning to a Compound Microscope

The single lens wasn't enough to resolve the image, so we upgrade our setup by adding an eyepiece, creating a compound microscope. The problem gives us two crucial pieces of information about this new setup:
1. The tube length $L$ is $0.6 	ext{ m}$, which we must convert to centimeters to keep our units consistent: $L = 60 	ext{ cm}$.
2. The final image is observed at an infinite distance.

For a compound microscope where the final image is formed at infinity, the magnifying power $M'$ is approximated by:

$$M' = \frac{L \cdot D}{f_0 \cdot f_e}$$

### Calculating the Eyepiece Focal Length

We are told that the new total magnification is double the earlier one. Since our initial magnification was 6, our new magnification $M'$ is:

$$M' = 2 	imes 6 = 12$$

Now, we have all the pieces of the puzzle. We know $M' = 12$, $L = 60 	ext{ cm}$, $D = 25 	ext{ cm}$, and we previously calculated $f_0 = 5 	ext{ cm}$. Let's substitute these into our compound microscope formula:

$$12 = \frac{60 	imes 25}{5 \cdot f_e}$$

Let's simplify the right side of the equation. Dividing 60 by 5 gives us 12:

$$12 = \frac{12 	imes 25}{f_e}$$

Notice how elegantly the math works out. The 12 on the left side perfectly cancels out the 12 on the right side in the numerator:

$$1 = \frac{25}{f_e}$$

Multiplying both sides by $f_e$, we arrive at our final answer:

$$f_e = 25 	ext{ cm}$$

**The required focal length of the eyepiece is 25 cm.** This problem perfectly illustrates how combining lenses allows us to push the boundaries of magnification!

Enter Numerical Value:

Visualized Solution

Visualizing the Simple Microscope

Magnification Formula for Simple Microscope

Substituting Values

Calculating Objective Focal Length

Visualizing the Compound Microscope

Magnification Formula for Compound Microscope

Substituting Values for Compound Microscope

Simplifying the Equation

Final Answer

The Way Forward

The Sigma Insight: Optical Instruments

The Simple Microscope Setup

Transitioning to a Compound Microscope

Calculating the Eyepiece Focal Length

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