The focal length of objective and the eyepiece of a compound microscope are 1cm and 5cm. An object is placed at a distance of 1.1 cm from the objective and if the final image is formed at infinity the magnifying power and distance between the lenses are respectively

Options

A. 60, 16cm
B. 50, 16cm
C. 50, 15.7cm
D. 60, 16.7cm

Answer

To solve the problem, we need to find two things: the magnifying power of the compound microscope and the distance between the lenses. Let's break it down step by step.

### Step 1: Identify Given Values
- Focal length of the objective lens, \( f_0 = 1 \, \text{cm} \)
- Focal length of the eyepiece lens, \( f_e = 5 \, \text{cm} \)
- Object distance from the objective lens, \( u_0 = -1.1 \, \text{cm} \) (negative because the object is on the same side as the incoming light)

### Step 2: Calculate Image Distance from the Objective Lens
Using the lens formula:
\[ \frac{1}{f_0} = \frac{1}{v_0} - \frac{1}{u_0} \]
Rearranging gives:
\[ \frac{1}{v_0} = \frac{1}{f_0} + \frac{1}{u_0} \]
Substituting the values:
\[ \frac{1}{v_0} = \frac{1}{1} + \frac{1}{-1.1} \]
Calculating:
\[ \frac{1}{v_0} = 1 - \frac{1}{1.1} = 1 - 0.909 = 0.091 \]
Thus,
\[ v_0 = \frac{1}{0.091} \approx 11 \, \text{cm} \]

### Step 3: Determine the Position of the Final Image
Since the final image is formed at infinity, the image formed by the objective lens must be at the focal point of the eyepiece lens. Therefore, the image distance from the eyepiece lens is equal to the focal length of the eyepiece:
\[ v_e = f_e = 5 \, \text{cm} \]

### Step 4: Calculate the Distance Between the Lenses
The distance between the lenses is the sum of the image distance from the objective lens and the focal length of the eyepiece:
\[ D = v_0 + f_e = 11 \, \text{cm} + 5 \, \text{cm} = 16 \, \text{cm} \]

### Step 5: Calculate the Magnifying Power
The magnifying power \( M \) of a compound microscope is given by:
\[ M = \frac{v_0}{u_0} \times \frac{D}{f_e} \]
Substituting the known values:
\[ M = \frac{11}{-1.1} \times \frac{16}{5} \]
Calculating:
\[ M = -10 \times 3.2 = -32 \]
The magnitude of magnification is:
\[ |M| = 32 \]

### Final Results
- Magnifying power: \( 32 \)
- Distance between the lenses: \( 16 \, \text{cm} \)

### Summary
The magnifying power of the compound microscope is \( 32 \) and the distance between the lenses is \( 16 \, \text{cm} \).

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