An object is placed beyond the centre of curvature $$C$$ of the given concave mirror. If the distance of the object is $$d_1$$ from $$C$$ and the distance of the image formed is $$d_2$$ from $$C$$, the radius of curvature of this mirror is:

Let us denote the pole of the concave mirror by $$P$$ and the centre of curvature by $$C$$. The radius of curvature is the distance $$PC$$, and we will write its magnitude as $$R$$.

The object is kept beyond the point $$C$$. Its distance from $$C$$ is given to be $$d_1$$, measured along the principal axis. Hence the distance of the object from the pole $$P$$ is the sum of the two segments $$PC$$ and $$CO$$:

$$PO = PC + CO = R + d_1.$$

According to the sign convention for spherical mirrors, all distances measured to the left of the pole are taken as negative. Therefore the object distance (symbol $$u$$) is

$$u = -(R + d_1).$$

The real image produced by a concave mirror, when the object is beyond $$C$$, lies between $$F$$ and $$C$$. Its distance from $$C$$ is given to be $$d_2$$. Since the image is to the left of the pole while remaining inside $$C$$, the distance of the image from the pole $$P$$ is

$$PI = PC - CI = R - d_2,$$

and, with the same sign convention, the image distance (symbol $$v$$) is

$$v = -(R - d_2).$$

Now we invoke the mirror formula, which relates the object distance $$u$$, the image distance $$v$$ and the focal length $$f$$:

$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}.$$

For a spherical mirror the focal length and the radius of curvature are connected by

$$f = \frac{R}{2}.$$

Because the mirror is concave (centre of curvature on the same side as the object), the focal length is negative, so

$$\frac{1}{f} = \frac{1}{-\dfrac{R}{2}} = -\frac{2}{R}.$$

Substituting $$u = -(R + d_1)$$, $$v = -(R - d_2)$$ and $$\dfrac{1}{f} = -\dfrac{2}{R}$$ in the mirror formula, we obtain

$$\frac{1}{-(R - d_2)} + \frac{1}{-(R + d_1)} = -\frac{2}{R}.$$

Removing the minus signs from the denominators gives

$$-\frac{1}{R - d_2} - \frac{1}{R + d_1} = -\frac{2}{R}.$$

Multiplying every term by $$-1$$ yields an equation with only positive numerators, easier to handle:

$$\frac{1}{R - d_2} + \frac{1}{R + d_1} = \frac{2}{R}.$$

We next add the two fractions on the left by writing them over a common denominator:

$$\frac{(R + d_1) + (R - d_2)}{(R + d_1)(R - d_2)} = \frac{2}{R}.$$

The numerator simplifies to

$$R + d_1 + R - d_2 = 2R + (d_1 - d_2).$$

Therefore the equation becomes

$$\frac{2R + (d_1 - d_2)}{(R + d_1)(R - d_2)} = \frac{2}{R}.$$

Cross-multiplying gets rid of the denominators:

$$R\bigl[2R + (d_1 - d_2)\bigr] = 2\,(R + d_1)(R - d_2).$$

Expanding each side separately, we have on the left

$$2R^2 + R(d_1 - d_2),$$

and on the right

$$2\left[R^2 - R\,d_2 + R\,d_1 - d_1 d_2\right] = 2R^2 - 2R d_2 + 2R d_1 - 2 d_1 d_2.$$

Now we bring everything to one side by subtracting the entire right-hand expression from the left-hand expression:

$$\bigl[2R^2 + R(d_1 - d_2)\bigr] - \bigl[2R^2 - 2R d_2 + 2R d_1 - 2 d_1 d_2\bigr] = 0.$$

The terms $$2R^2$$ cancel straight away, leaving

$$R(d_1 - d_2) + 2R d_2 - 2R d_1 + 2 d_1 d_2 = 0.$$

Collecting the coefficients of $$R$$ gives

$$R\bigl[(d_1 - d_2) + 2d_2 - 2d_1\bigr] + 2 d_1 d_2 = 0,$$

which simplifies to

$$R\bigl[-d_1 + d_2\bigr] + 2 d_1 d_2 = 0.$$

Rewriting the same relation in a slightly neater form,

$$R(d_2 - d_1) = -2 d_1 d_2.$$

Dividing both sides by $$d_2 - d_1$$ (note that for an object beyond $$C$$, $$d_1 > d_2$$, so the denominator is negative and the two minus signs cancel) we finally arrive at

$$R = \frac{2\,d_1\,d_2}{d_1 - d_2}.$$

This matches the expression given in option B.

Hence, the correct answer is Option B.