The focal lengths of the objective and eyelens of a microscope are 1.6 cm and 2.5 cm respectively. The distance between the two lenses is 21.7 cm. If the final image is formed at infinity, the distance between the object and the objective lens is :

Options

A. 1.8 cm
B. 1-70 cm
C. 1.65 cm
D. 1.75 cm

Explanation

To solve the problem step by step, we will use the given information about the focal lengths of the objective and eyepiece lenses, as well as the distance between the two lenses.

### Step 1: Identify the given values
- Focal length of the objective lens (F₀) = 1.6 cm
- Focal length of the eyepiece lens (Fₑ) = 2.5 cm
- Distance between the two lenses (L) = 21.7 cm

### Step 2: Use the formula for the length of the microscope
The length of the microscope (L) can be expressed as:
\[ L = V₀ + Fₑ \]
Where:
- V₀ = image distance from the objective lens

Substituting the known values:
\[ 21.7 = V₀ + 2.5 \]

### Step 3: Solve for V₀
Rearranging the equation to find V₀:
\[ V₀ = 21.7 - 2.5 \]
\[ V₀ = 19.2 \, \text{cm} \]

### Step 4: Apply the lens formula for the objective lens
The lens formula is given by:
\[ \frac{1}{V₀} - \frac{1}{U₀} = \frac{1}{F₀} \]
Where:
- U₀ = object distance from the objective lens

Substituting the known values into the lens formula:
\[ \frac{1}{19.2} - \frac{1}{U₀} = \frac{1}{1.6} \]

### Step 5: Rearrange the equation to find U₀
Rearranging gives us:
\[ \frac{1}{U₀} = \frac{1}{19.2} - \frac{1}{1.6} \]

### Step 6: Calculate the right-hand side
First, calculate \( \frac{1}{19.2} \) and \( \frac{1}{1.6} \):
\[ \frac{1}{19.2} \approx 0.05208 \]
\[ \frac{1}{1.6} = 0.625 \]

Now substituting these values:
\[ \frac{1}{U₀} = 0.05208 - 0.625 \]
\[ \frac{1}{U₀} = -0.57292 \]

### Step 7: Find U₀
Taking the reciprocal gives:
\[ U₀ = \frac{1}{-0.57292} \approx -1.75 \, \text{cm} \]

Since distance is typically taken as positive in this context, we take the absolute value:
\[ U₀ \approx 1.75 \, \text{cm} \]

### Conclusion
The distance between the object and the objective lens is approximately **1.75 cm**.