Animated Solution for Physics - Optics: 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

We are given an object placed beyond the center of curvature $C$. From our basic ray optics intuition, we know that when an object is placed beyond $C$, its real and inverted image forms between the center of curvature $C$ and the principal focus $F$.

The problem gives us distances relative to the center of curvature $C$: the object is at a distance $d_1$ from $C$, and the image is at a distance $d_2$ from $C$. To use our standard mirror formula, we must translate these relative distances into absolute coordinates measured from the pole $P$.

According to the Cartesian sign convention, since the object is placed in front of the reflecting surface, all our measurements will be negative.

The object distance $u$ is the distance from $P$ to $C$ plus the extra distance $d_1$. Thus, $u=-(R+d_1)$.

The image distance $v$ is the distance from $P$ to $C$ minus the distance $d_2$ (since the image is between $C$ and $F$). Thus, $v=-(R-d_2)$.

The focal length of a concave mirror is half its radius of curvature, so $f=-\frac{R}{2}$.

We start with the classic mirror formula:

To make the algebra cleaner, let's rearrange it to avoid dealing with reciprocals immediately:

Now, we substitute our carefully crafted expressions into this rearranged formula:

The negative signs on the left side multiply to give a positive product. On the right side, we can factor out the negative sign from the bracket:

Let's multiply the entire equation by 2 to eliminate the fraction, and expand the brackets:

Notice how the $2R^2$ terms on both sides perfectly cancel each other out! This is the beauty of physics equations; they often collapse into elegant forms.

Let's group the remaining terms containing $R$ on the left side, and move the $d_1{d}_2$ term to the right:

Finally, isolating $R$, we get our answer:

Pro Tip: This entire problem can be solved in just two lines using Newton's Formula, $x_1{x}_2=f^2$, where $x_1$ and $x_2$ are the distances of the object and image from the focus $F$. Try it out yourself to see how powerful alternative frameworks can be!