The length of the compound microscope is `14 cm.` The magnifying power for relaxed eye is `25`. If the focal length of eye lens is `5 cm`, then the object distance for objective lens will be

Options

A. `1.8cm`
B. `1.5 cm`
C. `2.1 cm`
D. `2.4 cm`

Explanation

To solve the problem step by step, we will use the given data and the formulas related to a compound microscope.

### Given Data:
- Length of the compound microscope (L) = 14 cm
- Magnifying power (M) = 25 (for relaxed eye)
- Focal length of the eye lens (Fe) = 5 cm

### Step 1: Understand the relationship in the compound microscope
In a compound microscope, the total length (L) is the distance from the objective lens to the eye lens. The image formed by the objective lens is at a distance (Vo) from the objective lens, and the distance from the eye lens to the final image (which is at infinity for a relaxed eye) is equal to the focal length of the eye lens (Fe).

### Step 2: Set up the equation for the length of the microscope
The length of the microscope can be expressed as:
\[ L = V_o + F_e \]
Where:
- \( V_o \) = distance from the objective lens to the image formed by the objective lens
- \( F_e \) = focal length of the eye lens

### Step 3: Substitute the known values
We know:
- \( L = 14 \) cm
- \( F_e = 5 \) cm

Substituting these values into the equation:
\[ 14 = V_o + 5 \]

### Step 4: Solve for \( V_o \)
Rearranging the equation to find \( V_o \):
\[ V_o = 14 - 5 \]
\[ V_o = 9 \text{ cm} \]

### Step 5: Use the magnifying power formula
The magnifying power (M) of a compound microscope is given by:
\[ M = \frac{V_o}{U_o} \cdot \left(1 + \frac{D}{F_e}\right) \]
Where:
- \( U_o \) = object distance from the objective lens
- \( D \) = least distance of distinct vision (approximately 25 cm)

### Step 6: Substitute the known values into the magnifying power formula
We know:
- \( M = 25 \)
- \( V_o = 9 \) cm
- \( D = 25 \) cm
- \( F_e = 5 \) cm

Substituting these values into the magnifying power formula:
\[ 25 = \frac{9}{U_o} \cdot \left(1 + \frac{25}{5}\right) \]

### Step 7: Simplify the equation
Calculating the term inside the parentheses:
\[ 1 + \frac{25}{5} = 1 + 5 = 6 \]

So, the equation becomes:
\[ 25 = \frac{9}{U_o} \cdot 6 \]

### Step 8: Solve for \( U_o \)
Rearranging to find \( U_o \):
\[ U_o = \frac{9 \cdot 6}{25} \]
\[ U_o = \frac{54}{25} \]
\[ U_o = 2.16 \text{ cm} \]

Thus, the object distance for the objective lens \( U_o \) is approximately 2.16 cm.

### Final Answer:
The object distance for the objective lens is \( U_o = 2.16 \text{ cm} \).

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