An object is to be seen through a simple microscope of power `10 D`. Where should an object be placed to produce maximum angular magnification ? Least distance of distinct vision is `25 cm`.

Explanation

To solve the problem of where to place an object to produce maximum angular magnification using a simple microscope with a power of \(10 D\), we can follow these steps:

### Step 1: Understand the relationship between power and focal length
The power \(P\) of a lens is related to its focal length \(f\) by the formula:
\[ P = \frac{100}{f} \quad \text{(in cm)} \]
Given that the power \(P = 10 D\), we can calculate the focal length \(f\):
\[ f = \frac{100}{P} = \frac{100}{10} = 10 \text{ cm} \]

**Hint:** Remember that the focal length is inversely proportional to the power of the lens.

### Step 2: Identify the condition for maximum angular magnification
For maximum angular magnification, the image should be formed at the least distance of distinct vision, which is given as \(D = 25 \text{ cm}\). Since the image is formed on the same side as the object for a simple microscope, the image distance \(v\) will be negative:
\[ v = -25 \text{ cm} \]

**Hint:** The least distance of distinct vision is the distance at which the eye can comfortably see the image.

### Step 3: Use the lens formula to find the object distance
The lens formula relates the object distance \(u\), image distance \(v\), and focal length \(f\):
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Rearranging this gives:
\[ \frac{1}{u} = \frac{1}{v} - \frac{1}{f} \]

Substituting the known values:
\[ v = -25 \text{ cm}, \quad f = 10 \text{ cm} \]
We have:
\[ \frac{1}{u} = \frac{1}{-25} - \frac{1}{10} \]

**Hint:** Make sure to keep track of the signs when substituting values into the lens formula.

### Step 4: Calculate \( \frac{1}{u} \)
Calculating the right-hand side:
\[ \frac{1}{u} = -\frac{1}{25} - \frac{1}{10} \]
Finding a common denominator (which is 50):
\[ \frac{1}{u} = -\frac{2}{50} - \frac{5}{50} = -\frac{7}{50} \]

### Step 5: Solve for \(u\)
Taking the reciprocal gives:
\[ u = -\frac{50}{7} \approx -7.14 \text{ cm} \]

The negative sign indicates that the object is placed on the same side as the incoming light, which is consistent with the conventions used in optics.

### Final Answer:
The object should be placed approximately \(7.14 \text{ cm}\) from the lens.

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