The focal lengths of the objective and eye`-` lens of a microscope are `1cm` and `5 cm` respectively. If the magnifying power for the relaxed eye is `45`, then the length of the tube is

Options

A. 30 cm
B. 25 cm
C. 15 cm
D. 12 cm

Explanation

To find the length of the tube of a microscope given the focal lengths of the objective and eyepiece lenses and the magnifying power, we can follow these steps:

### Step 1: Understand the given values
- Focal length of the objective lens, \( f_o = 1 \, \text{cm} \)
- Focal length of the eyepiece lens, \( f_e = 5 \, \text{cm} \)
- Magnifying power for the relaxed eye, \( M = 45 \)

### Step 2: Write the formula for magnifying power
The magnifying power \( M \) of a microscope can be expressed as:
\[ M = \frac{L - f_o}{f_e} \cdot \frac{D}{f_o} \]
where:
- \( L \) is the length of the tube,
- \( D \) is the distance of distinct vision, which is typically taken as \( 25 \, \text{cm} \).

### Step 3: Substitute the known values into the formula
Substituting the values we have:
\[ 45 = \frac{L - 1}{5} \cdot \frac{25}{1} \]

### Step 4: Simplify the equation
This simplifies to:
\[ 45 = \frac{25(L - 1)}{5} \]
\[ 45 = 5(L - 1) \]

### Step 5: Solve for \( L \)
Now, divide both sides by 5:
\[ 9 = L - 1 \]
Adding 1 to both sides gives:
\[ L = 10 \, \text{cm} \]

### Step 6: Calculate the total length of the tube
We must also consider the focal length of the eyepiece in the total length, so we add \( f_e \):
\[ L_{\text{total}} = L + f_e = 10 + 5 = 15 \, \text{cm} \]

### Final Answer
The length of the tube is \( 15 \, \text{cm} \).

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