A microscope is having objective of focal length and eye piece of focal length 6cm. If tube length 30cm and image is formed at the least distance of distant vision, what is the magnification prodcut by the microscope. (take D=25cm)

Options

A. 6
B. 150
C. 25
D. 125

Explanation

To solve the problem of finding the magnification produced by the microscope, we will follow these steps:

### Step 1: Identify the given values
- Focal length of the objective lens (\(f_o\)) = unknown (not provided in the question)
- Focal length of the eyepiece (\(f_e\)) = 6 cm
- Tube length (\(L\)) = 30 cm
- Least distance of distinct vision (\(D\)) = 25 cm

### Step 2: Write the formula for the total magnification of the microscope
The total magnification (\(M\)) produced by a microscope can be calculated using the formula:
\[ M = \left(\frac{L}{f_o}\right) \left(1 + \frac{D}{f_e}\right) \]

### Step 3: Substitute the known values into the formula
Since the focal length of the objective lens (\(f_o\)) is not provided, we will denote it as \(f_o\). We can now substitute the known values into the formula:
\[ M = \left(\frac{30}{f_o}\right) \left(1 + \frac{25}{6}\right) \]

### Step 4: Calculate the term inside the parentheses
First, calculate \(1 + \frac{25}{6}\):
\[ 1 + \frac{25}{6} = \frac{6}{6} + \frac{25}{6} = \frac{31}{6} \]

### Step 5: Substitute this back into the magnification formula
Now, substitute this back into the magnification formula:
\[ M = \left(\frac{30}{f_o}\right) \left(\frac{31}{6}\right) \]

### Step 6: Simplify the expression
Now simplify the expression:
\[ M = \frac{30 \times 31}{6 \times f_o} \]
\[ M = \frac{930}{6 f_o} \]
\[ M = \frac{155}{f_o} \]

### Step 7: Conclusion
The magnification produced by the microscope is given by:
\[ M = \frac{155}{f_o} \]
Since the value of \(f_o\) is not provided, we cannot calculate a numerical value for \(M\). However, the relationship shows that the magnification depends inversely on the focal length of the objective lens.