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

Explanation

To find the object distance for the objective lens in a compound microscope, we can follow these steps:

### Step 1: Understand the given data
- Length of the compound microscope (L) = 14 cm
- Magnifying power (M) = 25
- Focal length of the eyepiece (f_e) = 5 cm

### Step 2: Use the formula for magnifying power
The magnifying power (M) of a compound microscope is given by the formula:
\[ M = \frac{V_o}{U_o} \]
where \( V_o \) is the image distance from the objective lens and \( U_o \) is the object distance from the objective lens.

### Step 3: Relate magnifying power to the focal length of the eyepiece
Since we know that the magnifying power is also related to the focal length of the eyepiece, we can express it as:
\[ M = \frac{V_o}{U_o} = \frac{f_e}{f_o} \]
However, we will use the first relation for our calculations.

### Step 4: Relate the length of the microscope to distances
The length of the compound microscope is given by the relation:
\[ L = V_o + f_e \]
Substituting the known values:
\[ 14 = V_o + 5 \]

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

### Step 6: Substitute \( V_o \) back into the magnifying power formula
Now, we can substitute \( V_o \) into the magnifying power formula:
\[ M = \frac{V_o}{U_o} \]
Substituting \( M = 25 \) and \( V_o = 9 \):
\[ 25 = \frac{9}{U_o} \]

### Step 7: Solve for \( U_o \)
Rearranging gives:
\[ U_o = \frac{9}{25} \]
Calculating this gives:
\[ U_o = 0.36 \text{ cm} \]

### Conclusion
The object distance for the objective lens \( U_o \) is \( 0.36 \text{ cm} \).

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