The magnifying powerof a compound microscope is 20 and the distance between its two lenses is 30 cm when the final image is at the near point of the eye If the focal length of eye-pfece is 6.25 cm, the focal length of objective is

Options

A. 2.5 cm
B. 3.5 cm
C. 4.5 cm
D. 5.0 cm

Explanation

To solve the problem, we need to find the focal length of the objective lens of a compound microscope given the magnifying power, the distance between the lenses, and the focal length of the eyepiece.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- Magnifying power (M) of the compound microscope = 20
- Distance between the two lenses (L) = 30 cm
- Focal length of the eyepiece (Fe) = 6.25 cm
- The final image is at the near point of the eye (D = 25 cm).

2. **Formula for Magnifying Power**:
The magnifying power of a compound microscope is given by the product of the magnifying power of the objective lens (Mo) and the eyepiece (Me):
\[ M = Mo \times Me \]

3. **Finding the Magnifying Power of the Eyepiece (Me)**:
The magnifying power of the eyepiece can be calculated using the formula:
\[ Me = 1 + \frac{D}{Fe} \]
Substituting the values:
\[ Me = 1 + \frac{25}{6.25} = 1 + 4 = 5 \]

4. **Finding the Magnifying Power of the Objective Lens (Mo)**:
Now, we can find the magnifying power of the objective lens:
\[ M = Mo \times Me \implies 20 = Mo \times 5 \implies Mo = \frac{20}{5} = 4 \]

5. **Using the Magnification Formula for the Objective Lens**:
The magnifying power of the objective lens is also given by:
\[ Mo = \frac{V_o}{U_o} \]
Where \( V_o \) is the image distance and \( U_o \) is the object distance for the objective lens.

6. **Finding the Image Distance (Vo)**:
The distance between the two lenses is given as 30 cm. If \( U_e \) is the object distance for the eyepiece, then:
\[ V_o = L - U_e \]
First, we need to find \( U_e \) using the lens formula for the eyepiece:
\[ \frac{1}{V_e} - \frac{1}{U_e} = \frac{1}{Fe} \]
Here, \( V_e = -25 \) cm (the image formed by the eyepiece is virtual):
\[ \frac{1}{-25} - \frac{1}{U_e} = \frac{1}{6.25} \]
Rearranging gives:
\[ -\frac{1}{U_e} = \frac{1}{6.25} + \frac{1}{25} \]
Finding a common denominator (which is 25):
\[ -\frac{1}{U_e} = \frac{4 + 1}{25} = \frac{5}{25} = \frac{1}{5} \]
Therefore,
\[ U_e = -5 \text{ cm} \]

7. **Calculating the Image Distance (Vo)**:
Now substituting \( U_e \) back into the equation for \( V_o \):
\[ V_o = 30 - (-5) = 30 + 5 = 35 \text{ cm} \]

8. **Finding the Object Distance (Uo)**:
Now we can find \( U_o \):
\[ Mo = \frac{V_o}{U_o} \implies 4 = \frac{35}{U_o} \implies U_o = \frac{35}{4} = 8.75 \text{ cm} \]

9. **Finding the Focal Length of the Objective Lens (Fo)**:
We can now use the lens formula for the objective lens:
\[ \frac{1}{V_o} - \frac{1}{U_o} = \frac{1}{F_o} \]
Substituting the values:
\[ \frac{1}{35} - \frac{1}{8.75} = \frac{1}{F_o} \]
Finding a common denominator (which is 35 * 8.75):
\[ \frac{8.75 - 35}{35 \times 8.75} = \frac{1}{F_o} \]
Simplifying gives:
\[ F_o = 5 \text{ cm} \]

### Final Answer:
The focal length of the objective lens is **5 cm**.