In a compound microscope, the focal lengths of two lenses are `1.5 cm` and `6.25 cm` an object is placed at `2 cm` form objective and the final image is formed at `25 cm` from eye lens. The distance between the two lenses is

Options

A. `6.00 cm`
B. `7.75 cm`
C. `9.25 cm`
D. `11.0 cm`

Explanation

To find the distance between the two lenses of a compound microscope, we will follow these steps:

### Step 1: Identify the given values
- Focal length of the objective lens, \( F_O = 1.5 \, \text{cm} \)
- Focal length of the eye lens, \( F_E = 6.25 \, \text{cm} \)
- Object distance from the objective lens, \( U_O = -2 \, \text{cm} \) (negative as per sign convention)
- Final image distance from the eye lens, \( V_E = 25 \, \text{cm} \)

### Step 2: Use the lens formula for the objective lens
The lens formula is given by:
\[ \frac{1}{V_O} - \frac{1}{U_O} = \frac{1}{F_O} \]
Substituting the known values:
\[ \frac{1}{V_O} - \frac{1}{-2} = \frac{1}{1.5} \]
This simplifies to:
\[ \frac{1}{V_O} + \frac{1}{2} = \frac{2}{3} \]

### Step 3: Solve for \( V_O \)
Rearranging the equation gives:
\[ \frac{1}{V_O} = \frac{2}{3} - \frac{1}{2} \]
Finding a common denominator (6):
\[ \frac{1}{V_O} = \frac{4}{6} - \frac{3}{6} = \frac{1}{6} \]
Thus,
\[ V_O = 6 \, \text{cm} \]

### Step 4: Use the tube length formula for the microscope
The tube length \( L \) is given by:
\[ L = V_O + \frac{D \cdot F_E}{D + F_E} \]
Where \( D = V_E \) (the distance of the final image from the eye lens). Substituting the values:
\[ L = 6 + \frac{25 \cdot 6.25}{25 + 6.25} \]

### Step 5: Calculate the distance between the lenses
Calculating the denominator:
\[ 25 + 6.25 = 31.25 \]
Now substituting back into the equation:
\[ L = 6 + \frac{156.25}{31.25} \]
Calculating \( \frac{156.25}{31.25} \):
\[ \frac{156.25}{31.25} = 5 \]
So,
\[ L = 6 + 5 = 11 \, \text{cm} \]

### Final Answer
The distance between the two lenses is \( 11 \, \text{cm} \).

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