An eye can distinguish between two points of an object if they are separated by more than 0.22 mm when the object is placed at 25 cm from the eye. The object is now seen by a compound microscope having a 20 D objective and 10 D eyepiece separated by a distance of 20 cm. The final image is formed at 25 cm from the eye. What is the minimum separation between two points of the object which can now be distinguished?

Explanation

To solve the problem, we need to find the minimum separation between two points of the object that can now be distinguished when viewed through a compound microscope. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the given data
- The minimum separation between two points of an object distinguishable by the eye is \( d = 0.22 \, \text{mm} = 0.022 \, \text{cm} \).
- The object is viewed through a compound microscope with:
- Objective lens power \( P_o = 20 \, \text{D} \)
- Eyepiece lens power \( P_e = 10 \, \text{D} \)
- Distance between the lenses \( D = 20 \, \text{cm} \)
- The final image is formed at a distance of \( 25 \, \text{cm} \) from the eye.

### Step 2: Calculate the focal lengths of the lenses
The focal length \( f \) of a lens can be calculated using the formula:
\[ f = \frac{100}{P} \, \text{cm} \]
- For the objective lens:
\[ f_o = \frac{100}{20} = 5 \, \text{cm} \]
- For the eyepiece lens:
\[ f_e = \frac{100}{10} = 10 \, \text{cm} \]

### Step 3: Apply the lens formula for the eyepiece
Using the lens formula:
\[ \frac{1}{f_e} = \frac{1}{V_e} - \frac{1}{U_e} \]
Where:
- \( V_e = -25 \, \text{cm} \) (image distance for the eyepiece, negative because it is virtual)
- \( U_e \) is the object distance for the eyepiece.

Substituting the values:
\[ \frac{1}{10} = -\frac{1}{25} - \frac{1}{U_e} \]
Rearranging gives:
\[ \frac{1}{U_e} = -\frac{1}{25} - \frac{1}{10} \]
Calculating the right side:
\[ \frac{1}{U_e} = -\frac{2 + 5}{50} = -\frac{7}{50} \]
Thus:
\[ U_e = -\frac{50}{7} \approx -7.14 \, \text{cm} \]

### Step 4: Calculate the object distance for the objective lens
The total distance between the two lenses is \( D = 20 \, \text{cm} \):
\[ U_o = D - U_e = 20 - \left(-\frac{50}{7}\right) = 20 + \frac{50}{7} = \frac{140 + 50}{7} = \frac{190}{7} \approx 27.14 \, \text{cm} \]

### Step 5: Apply the lens formula for the objective lens
Using the lens formula for the objective lens:
\[ \frac{1}{f_o} = \frac{1}{V_o} - \frac{1}{U_o} \]
Where \( V_o \) is the image distance from the objective lens. We can rearrange this to find \( V_o \).

### Step 6: Calculate magnification
The magnification \( M \) of the compound microscope is given by:
\[ M = -\frac{V_o}{U_o} \left(1 + \frac{D}{f_e}\right) \]
Substituting the values we have calculated.

### Step 7: Calculate the new minimum separation
The new minimum separation \( d' \) that can be distinguished through the microscope is:
\[ d' = \frac{d}{M} \]
Where \( d \) is the original minimum separation.

### Final Calculation
Substituting the values into the equations will yield the final answer.

### Final Answer
The minimum separation between two points of the object which can now be distinguished is approximately \( 0.04 \, \text{mm} \).

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