A compound microscope consists of an objective lens with focal length `1.0 cm` and eye piece of the focal length `2.0` cm and a tube `20 cm` from eye lens. Now final image is formed at the near point of eye, so find the magnifying power of microscope.

Options

A. `300 cm`
B. `232 cm`
C. `290 cm`
D. `270 cm`

Explanation

To find the magnifying power of the compound microscope, we will follow these steps:

### Step 1: Understand the Components
We have:
- Focal length of the objective lens, \( f_o = 1.0 \, \text{cm} \)
- Focal length of the eyepiece lens, \( f_e = 2.0 \, \text{cm} \)
- Tube length, \( L = 20 \, \text{cm} \)
- Final image formed at the near point, \( D = 25 \, \text{cm} \)

### Step 2: Calculate the Image Distance from the Eyepiece
The image formed by the objective lens acts as an object for the eyepiece. The distance of the image from the eyepiece, \( v_e \), can be found using the formula:
\[ v_e = D = 25 \, \text{cm} \]

Using the lens formula:
\[ \frac{1}{f_e} = \frac{1}{v_e} - \frac{1}{u_e} \]
Rearranging gives:
\[ \frac{1}{u_e} = \frac{1}{v_e} - \frac{1}{f_e} \]
Substituting the values:
\[ \frac{1}{u_e} = \frac{1}{25} - \frac{1}{2} \]
Calculating:
\[ \frac{1}{u_e} = \frac{1 - 12.5}{25} = \frac{-11.5}{25} \]
Thus,
\[ u_e = -\frac{25}{11.5} \approx -2.17 \, \text{cm} \]

### Step 3: Calculate the Image Distance from the Objective Lens
Using the tube length \( L \):
\[ v_o = L - u_e \]
Substituting the values:
\[ v_o = 20 - (-2.17) = 20 + 2.17 = 22.17 \, \text{cm} \]

### Step 4: Calculate the Object Distance for the Objective Lens
Using the lens formula for the objective lens:
\[ \frac{1}{f_o} = \frac{1}{v_o} - \frac{1}{u_o} \]
Rearranging gives:
\[ \frac{1}{u_o} = \frac{1}{v_o} - \frac{1}{f_o} \]
Substituting the values:
\[ \frac{1}{u_o} = \frac{1}{22.17} - \frac{1}{1} \]
Calculating:
\[ \frac{1}{u_o} = \frac{1 - 22.17}{22.17} = \frac{-21.17}{22.17} \]
Thus,
\[ u_o = -\frac{22.17}{21.17} \approx -1.05 \, \text{cm} \]

### Step 5: Calculate the Magnifying Power
The magnifying power \( M \) of the compound microscope is given by:
\[ M = M_o \times M_e \]
Where:
- \( M_o = \frac{v_o}{u_o} \) (magnification by the objective)
- \( M_e = 1 + \frac{D}{f_e} \) (magnification by the eyepiece)

Calculating \( M_o \):
\[ M_o = \frac{22.17}{-1.05} \approx -21.1 \]
Calculating \( M_e \):
\[ M_e = 1 + \frac{25}{2} = 1 + 12.5 = 13.5 \]
Thus, the total magnifying power:
\[ M = -21.1 \times 13.5 \approx -284.85 \]

### Final Answer
The magnifying power of the compound microscope is approximately \( 284.85 \).

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