When an object is viewed with a light of wavelength `6000Å` under a microscope, its resolving power is `10^(4)`. The resolving power of the microscope when the same object is viewed with a light of wavelength `4000Å`, is `nxx10^(3)`. The vlaue of n is

Answer

To solve the problem, we need to find the resolving power of a microscope when viewing an object with a light of wavelength `4000Å`, given that the resolving power with a wavelength of `6000Å` is `10^4`.

### Step-by-Step Solution:

1. **Understanding Resolving Power**:
The resolving power (R) of a microscope is inversely proportional to the wavelength (λ) of the light used. This can be expressed as:
\[ R \propto \frac{1}{\lambda} \]

2. **Setting Up the Proportionality**:
For two different wavelengths, we can write:
\[ \frac{R_A}{R_B} = \frac{\lambda_B}{\lambda_A} \]
where:
- \( R_A \) is the resolving power with wavelength \( \lambda_A = 4000Å \)
- \( R_B \) is the resolving power with wavelength \( \lambda_B = 6000Å \)

3. **Substituting Known Values**:
We know:
- \( R_B = 10^4 \)
- \( \lambda_A = 4000Å \)
- \( \lambda_B = 6000Å \)

Substituting these values into the equation gives:
\[ \frac{R_A}{10^4} = \frac{6000}{4000} \]

4. **Calculating the Ratio**:
Simplifying the right side:
\[ \frac{6000}{4000} = \frac{3}{2} \]
Therefore, we have:
\[ \frac{R_A}{10^4} = \frac{3}{2} \]

5. **Finding \( R_A \)**:
Rearranging the equation to find \( R_A \):
\[ R_A = 10^4 \times \frac{3}{2} \]

6. **Calculating \( R_A \)**:
\[ R_A = 10^4 \times 1.5 = 1.5 \times 10^4 \]

7. **Expressing \( R_A \) in Required Form**:
We need to express \( R_A \) in the form \( n \times 10^3 \):
\[ R_A = 15 \times 10^3 \]

8. **Finding \( n \)**:
From the expression \( R_A = n \times 10^3 \), we can see that:
\[ n = 15 \]

### Final Answer:
The value of \( n \) is **15**.

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