Animated Solution for Physics - Optics: A spherical mirror is obtained as shown in the above figure from a hollow glass sphere. If an object is in front of the mirror, what will be the nature and magnification of the image of the object? (Figure drawn as schematic and not to scale)

The journey to solving this optics problem begins with a careful observation of the provided diagram. At first glance, it might look like a simple spherical mirror, but the scale holds the key to unlocking the entire solution. Let's break it down step-by-step.

When we look at the diagram, we see a mirror cut from a hollow glass sphere. The dashed lines represent the original sphere, and the solid arc is our mirror. Notice the small lines on the outer surface? That indicates silvering. Since the outer surface is silvered, the inner surface is the reflecting one, making this a concave mirror.

Next, we need to extract the numerical parameters from the scale. The pole of the mirror is at the origin ($0\text{ cm}$). The center of the sphere, which corresponds to the center of curvature $C$, aligns with the $8\text{ cm}$ mark on the scale. According to the Cartesian sign convention, since the center lies to the left of the pole (against the direction of incident light), the radius of curvature is $R=-8\text{ cm}$.

The focal length $f$ is exactly half of the radius of curvature:

Now, where is the object? The object arrow is placed exactly halfway between the $8\text{ cm}$ and $12\text{ cm}$ marks. This means the object distance is $u=-10\text{ cm}$.

Mirror Formula

With our parameters locked in, we can determine exactly where the image will form. We use the fundamental mirror formula:

Substituting our known values into the equation:

Let's isolate $\frac{1}{v}$ by moving the object distance term to the right side:

To add these fractions, we find a common denominator, which is $20$:

Flipping the fraction gives us the image distance:

The negative sign tells us that the image is formed in front of the mirror, which means it is a real image.

Knowing the position is great, but we also need to know the size and orientation of the image. For this, we use the magnification formula:

Plugging in our values for $v$ and $u$:

The negative signs in the numerator and denominator cancel out, leaving us with:

The magnification value $m=-2/3$ gives us the final pieces of the puzzle.

1. The negative sign indicates that the image is formed below the principal axis, meaning it is inverted. (And as a rule of thumb for single mirrors, inverted images are always real). 2. The magnitude $|m|=2/3$ is less than $1$, which means the image is smaller than the object. Therefore, it is diminished or unmagnified.

Combining all our findings, the image is inverted, real, and unmagnified. This perfectly matches option (d).

Pro Tip: You could have also solved this conceptually! The object is at $10\text{ cm}$, which is between the center of curvature ($8\text{ cm}$) and infinity. For a concave mirror, any object placed beyond $C$ will always produce an image between $C$ and $F$ that is real, inverted, and diminished. Visualizing the ray diagram can save you precious time in an exam!