The diameter of the objective lens of the microscope makes an angle $$\beta$$ at the focus of the microscope. Further, the medium between the object and the lens is the oil of the refractive index n. Then the resolving power of the microscope:

The resolving power of a microscope is its ability to distinguish two closely spaced objects as separate. It is given by the formula:

$$ \text{RP} = \frac{2n \sin \theta}{\lambda} $$

where $$ n $$ is the refractive index of the medium, $$ \theta $$ is the semi-vertical angle (the angle between the axis of the lens and a ray from the edge of the object to the lens), and $$ \lambda $$ is the wavelength of light used.

In this problem, the angle $$ \beta $$ is defined as the angle subtended by the diameter of the objective lens at the focus. This angle $$ \beta $$ is the full angle, which is twice the semi-vertical angle $$ \theta $$. Therefore, we have:

$$ \beta = 2\theta \quad \text{or} \quad \theta = \frac{\beta}{2} $$

Substituting $$ \theta = \frac{\beta}{2} $$ into the resolving power formula:

$$ \text{RP} = \frac{2n \sin \left( \frac{\beta}{2} \right)}{\lambda} $$

For microscopes, the angles involved are typically small, so we can use the small-angle approximation. For small angles, $$ \sin \left( \frac{\beta}{2} \right) \approx \frac{\beta}{2} $$. Substituting this approximation:

$$ \text{RP} \approx \frac{2n \cdot \frac{\beta}{2}}{\lambda} = \frac{n \beta}{\lambda} $$

Now, consider the expression $$ n \sin(2\beta) $$. For small angles, $$ \sin(2\beta) \approx 2\beta $$, so:

$$ n \sin(2\beta) \approx n \cdot 2\beta = 2n\beta $$

From the approximation $$ \text{RP} \approx \frac{n \beta}{\lambda} $$, we can write:

$$ \text{RP} \approx \frac{n \beta}{\lambda} = \frac{1}{2\lambda} \cdot (2n\beta) \propto n \sin(2\beta) $$

since $$ 2n\beta \approx n \sin(2\beta) $$. Therefore, the resolving power is proportional to $$ n \sin(2\beta) $$. This means that as $$ n \sin(2\beta) $$ increases, the resolving power increases.

Now, evaluate the options:

Option A: Increases with decreasing value of $$ \beta $$. From $$ \text{RP} \approx \frac{n \beta}{\lambda} $$, decreasing $$ \beta $$ decreases RP, so this is false.

Option B: Increases with increasing value of $$ n \sin 2\beta $$. As derived, RP increases with increasing $$ n \sin(2\beta) $$, so this is true.

Option C: Increases with increasing value of $$ \frac{1}{n \sin 2\beta} $$. If $$ \frac{1}{n \sin 2\beta} $$ increases, then $$ n \sin 2\beta $$ decreases, which would decrease RP, so this is false.

Option D: Increases with decreasing value of $$ n $$. From $$ \text{RP} \approx \frac{n \beta}{\lambda} $$, decreasing $$ n $$ decreases RP, so this is false.

Hence, the correct answer is Option B.