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. 155
C. 25
D. 125

Answer

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 eyepiece (FE) = 6 cm
- Tube length (L) = 30 cm
- Least distance of distinct vision (D) = 25 cm
- Focal length of the objective (FO) = 1 cm (not explicitly stated, but inferred from the context)

### Step 2: Calculate the magnification of the objective (MO)
The formula for the magnification of the objective lens is given by:
\[ MO = -\frac{L}{FO} \]
Substituting the values:
\[ MO = -\frac{30 \text{ cm}}{1 \text{ cm}} = -30 \]

### Step 3: Calculate the magnification of the eyepiece (ME)
The formula for the magnification of the eyepiece is given by:
\[ ME = \frac{D}{FE} \]
Substituting the values:
\[ ME = \frac{25 \text{ cm}}{6 \text{ cm}} \approx 4.17 \]

### Step 4: Calculate the total magnification (M)
The total magnification of the microscope is the product of the magnifications of the objective and the eyepiece:
\[ M = ME \times MO \]
Substituting the values:
\[ M = 4.17 \times (-30) \approx -125 \]

### Step 5: Interpret the result
The negative sign indicates that the image is inverted. The magnitude of the magnification is 125.

### Final Answer
The magnification produced by the microscope is approximately 125.

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