An angular magnification (magnifying power) of `30 X` is desired using an objective of focal length `1.25 cm` and an eye piece of focal length `5 cm`. How will you set up the compound microscope ?

Explanation

To set up a compound microscope with an angular magnification of 30X using an objective lens with a focal length of 1.25 cm and an eyepiece with a focal length of 5 cm, we can follow these steps:

### Step 1: Understand the Magnification Formula
The total magnification (M) of a compound microscope is given by the product of the magnification produced by the objective (Mo) and the magnification produced by the eyepiece (Me):
\[ M = Mo \times Me \]
Given that \( M = 30 \), we need to find \( Mo \) and \( Me \).

### Step 2: Calculate the Magnification of the Eyepiece
The magnification of the eyepiece is given by:
\[ Me = 1 + \frac{D}{Fe} \]
where \( D \) is the least distance of distinct vision (approximately 25 cm) and \( Fe \) is the focal length of the eyepiece (5 cm).
Substituting the values:
\[ Me = 1 + \frac{25}{5} = 1 + 5 = 6 \]

### Step 3: Calculate the Required Magnification of the Objective
Now, we can find the required magnification of the objective:
\[ Mo = \frac{M}{Me} = \frac{30}{6} = 5 \]

### Step 4: Relate the Object and Image Distances for the Objective
The magnification of the objective 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. Since \( Mo = 5 \):
\[ 5 = -\frac{v_o}{u_o} \]
This implies:
\[ v_o = -5u_o \]

### Step 5: Use the Lens Formula for the Objective
The lens formula is given by:
\[ \frac{1}{f_o} = \frac{1}{v_o} - \frac{1}{u_o} \]
Substituting \( f_o = 1.25 \) cm:
\[ \frac{1}{1.25} = \frac{1}{v_o} - \frac{1}{u_o} \]

### Step 6: Substitute \( v_o \) in the Lens Formula
Substituting \( v_o = -5u_o \) into the lens formula:
\[ \frac{1}{1.25} = \frac{1}{-5u_o} - \frac{1}{u_o} \]
This simplifies to:
\[ \frac{1}{1.25} = -\frac{1 + 5}{5u_o} = -\frac{6}{5u_o} \]

### Step 7: Solve for \( u_o \)
Rearranging gives:
\[ u_o = -\frac{6 \times 1.25}{5} = -1.5 \text{ cm} \]

### Step 8: Calculate \( v_o \)
Now, substituting \( u_o \) back to find \( v_o \):
\[ v_o = -5(-1.5) = 7.5 \text{ cm} \]

### Step 9: Find the Separation Between the Objective and Eyepiece
The separation \( d \) between the objective and the eyepiece can be found by:
\[ d = v_o + |u_e| \]
where \( u_e \) is the object distance for the eyepiece. Since the image formed by the objective acts as the object for the eyepiece, we can find \( u_e \) using the eyepiece lens formula.

### Step 10: Calculate \( u_e \)
Using the eyepiece lens formula:
\[ \frac{1}{f_e} = \frac{1}{v_e} - \frac{1}{u_e} \]
Given \( f_e = 5 \) cm and \( v_e = -25 \) cm (for least distance of distinct vision):
\[ \frac{1}{5} = \frac{1}{-25} - \frac{1}{u_e} \]
Rearranging gives:
\[ \frac{1}{u_e} = -\frac{1}{25} - \frac{1}{5} = -\frac{1 + 5}{25} = -\frac{6}{25} \]
Thus:
\[ u_e = -\frac{25}{6} \approx -4.17 \text{ cm} \]

### Step 11: Calculate the Separation
Now substituting \( v_o \) and \( |u_e| \):
\[ d = 7.5 + 4.17 = 11.67 \text{ cm} \]

### Final Answer
The separation between the objective lens and the eyepiece lens should be approximately **11.67 cm**.