Two points separated by a distance of `0.1mm` can just be resolved in a microscope when a light of wavelength `6000Å` is used. If the light of wavelength `4800Å` is used this limit of resolution becomes

Options

A. 0.8 mm
B. 0.12 mm
C. 0.1 mm
D. 0.08 mm

Explanation

To solve the problem of finding the new limit of resolution when using light of a different wavelength, we can follow these steps:

### Step 1: Understand the formula for limit of resolution
The limit of resolution (d) for a microscope is given by the formula:
\[ d = \frac{1.22 \lambda}{D} \]
where:
- \( d \) is the limit of resolution,
- \( \lambda \) is the wavelength of light used,
- \( D \) is the diameter of the aperture (or pupil).

### Step 2: Identify the known values
From the problem:
- The initial distance that can be resolved (d) is \( 0.1 \, \text{mm} = 0.1 \times 10^{-3} \, \text{m} \).
- The initial wavelength (\( \lambda_1 \)) is \( 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} = 6 \times 10^{-7} \, \text{m} \).
- The new wavelength (\( \lambda_2 \)) is \( 4800 \, \text{Å} = 4800 \times 10^{-10} \, \text{m} = 4.8 \times 10^{-7} \, \text{m} \).
- The diameter (D) is constant and can be derived from the initial condition.

### Step 3: Calculate the diameter (D) using the initial condition
Using the initial condition:
\[ 0.1 \times 10^{-3} = \frac{1.22 \times 6 \times 10^{-7}}{D} \]
Rearranging gives:
\[ D = \frac{1.22 \times 6 \times 10^{-7}}{0.1 \times 10^{-3}} \]
Calculating this:
\[ D = \frac{1.22 \times 6 \times 10^{-7}}{0.1 \times 10^{-3}} = \frac{7.32 \times 10^{-7}}{0.1 \times 10^{-3}} = 7.32 \times 10^{-3} \, \text{m} = 7.32 \, \text{mm} \]

### Step 4: Calculate the new limit of resolution with the new wavelength
Now, substituting the new wavelength into the limit of resolution formula:
\[ d' = \frac{1.22 \times 4.8 \times 10^{-7}}{7.32 \times 10^{-3}} \]
Calculating this:
\[ d' = \frac{1.22 \times 4.8 \times 10^{-7}}{7.32 \times 10^{-3}} = \frac{5.856 \times 10^{-7}}{7.32 \times 10^{-3}} \]
\[ d' = 8.00 \times 10^{-5} \, \text{m} = 0.08 \, \text{mm} \]

### Step 5: Final result
Thus, the limit of resolution when using light of wavelength \( 4800 \, \text{Å} \) is \( 0.08 \, \text{mm} \).