A person with a normal near point `(25 cm)` using a compound microscope with an objective of focal length `8.0 mm` and eye piece of focal length `2.5 cm` can bring an object placed `9.0 cm` from the objective in sharp focus. What is the separation between the two lenses ? Calculate the magnifying power of the microscope ?

Options

A. `9.47 cm, 88`
B. `3.36 cm, 44`
C. `6.00 cm, 22`
D. `4.79 cm, 11`

Explanation

To solve the problem, we need to find the separation between the two lenses of the compound microscope and calculate its magnifying power. Let's break down the solution step by step.

### Step 1: Identify the Given Data
- Focal length of the objective lens, \( f_o = 8.0 \, \text{mm} = 0.8 \, \text{cm} \)
- Focal length of the eyepiece, \( f_e = 2.5 \, \text{cm} \)
- Object distance from the objective lens, \( u_o = -9.0 \, \text{cm} \) (negative because the object is on the same side as the incoming light)
- Near point (least distance of distinct vision), \( D = 25 \, \text{cm} \)

### Step 2: Calculate the Image Distance from the Objective Lens
Using the lens formula:
\[ \frac{1}{f_o} = \frac{1}{v_o} - \frac{1}{u_o} \]
Rearranging gives:
\[ \frac{1}{v_o} = \frac{1}{f_o} + \frac{1}{u_o} \]
Substituting the values:
\[ \frac{1}{v_o} = \frac{1}{0.8} + \frac{1}{-9.0} \]
Calculating:
\[ \frac{1}{v_o} = 1.25 - 0.1111 = 1.1389 \]
Thus,
\[ v_o = \frac{1}{1.1389} \approx 0.878 \, \text{cm} \approx 7.2 \, \text{cm} \]

### Step 3: Calculate the Image Distance from the Eyepiece
The image formed by the objective lens acts as the object for the eyepiece. The distance of this image from the eyepiece, \( u_e \), is given by:
\[ u_e = - (v_o - d) \]
where \( d \) is the separation between the two lenses. The image distance from the eyepiece, \( v_e \), is given as the near point distance:
\[ v_e = -D = -25 \, \text{cm} \]
Using the lens formula for the eyepiece:
\[ \frac{1}{f_e} = \frac{1}{v_e} - \frac{1}{u_e} \]
Rearranging gives:
\[ \frac{1}{u_e} = \frac{1}{v_e} - \frac{1}{f_e} \]
Substituting the values:
\[ \frac{1}{u_e} = \frac{1}{-25} - \frac{1}{2.5} \]
Calculating:
\[ \frac{1}{u_e} = -0.04 - 0.4 = -0.44 \]
Thus,
\[ u_e = \frac{1}{-0.44} \approx -2.27 \, \text{cm} \]

### Step 4: Calculate the Separation Between the Lenses
The separation \( L \) between the two lenses is given by:
\[ L = |u_e| + v_o \]
Substituting the values:
\[ L = |-2.27| + 7.2 = 2.27 + 7.2 = 9.47 \, \text{cm} \]

### Step 5: Calculate the Magnifying Power of the Microscope
The magnifying power \( M \) of the microscope is given by:
\[ M = \frac{v_o}{|u_o|} \times \left(1 + \frac{D}{f_e}\right) \]
Substituting the values:
\[ M = \frac{7.2}{9.0} \times \left(1 + \frac{25}{2.5}\right) \]
Calculating:
\[ M = 0.8 \times (1 + 10) = 0.8 \times 11 = 8.8 \approx 88 \]

### Final Answers
- The separation between the two lenses is \( 9.47 \, \text{cm} \).
- The magnifying power of the microscope is \( 88 \).