Q.The magnification produced by a spherical mirror is $-2.0$. The mirror used and the nature of the image formed will be: (A) Convex and virtual (B) Concave and real (C) Concave and virtual (D) Convex and real

Concept and Intuition

The magnification $m$ of a spherical mirror tells you two things at once: the sign tells you orientation, and the magnitude tells you size.

• If $m$ is positive, the image is virtual and erect (upright).
• If $m$ is negative, the image is real and inverted (upside down).

The magnitude $|m|$ tells you relative size:

• $|m| > 1$ → image is magnified (larger than object)
• $|m| < 1$ → image is diminished (smaller)
• $|m| = 1$ → same size

Here $m = -2.0$ means the image is inverted (negative) and twice as large as the object ($|m| = 2$).

Now, which mirror can produce an inverted, magnified image? A convex mirror always gives a virtual, erect, and diminished image — so it can never produce a negative magnification. A concave mirror, however, can produce both real (inverted) and virtual (erect) images depending on where the object is placed. The real image from a concave mirror is always inverted, and when the object is between the centre of curvature and the focus, that real image is also magnified.

So the only mirror that fits $m = -2.0$ is a concave mirror, and the image must be real.

Step-by-step reasoning

1. Interpret the sign of $m$

   $m = -2.0$ is negative. For spherical mirrors, a negative magnification always means the image is inverted relative to the object. An inverted image formed by a single mirror is always real (it can be projected on a screen). So the image is real.

2. Interpret the magnitude of $m$

   $|m| = 2.0 > 1$, so the image is magnified — larger than the object.

3. Eliminate convex mirror

   A convex mirror always produces a virtual, erect, and diminished image for any real object. That means $m$ is always positive and $|m| < 1$. Since our $m$ is negative and $|m| > 1$, a convex mirror is impossible. This eliminates options (A) and (D).

4. Check concave mirror possibilities

   A concave mirror can produce:

   • A real, inverted, magnified image when the object is placed between $F$ and $C$ (focus and centre of curvature).
   • A virtual, erect, magnified image when the object is placed between $P$ and $F$ (pole and focus). In that case $m$ is positive.

   Since our $m$ is negative, the image cannot be virtual. So the only possibility is the real, inverted, magnified case — which is exactly what a concave mirror gives for an object between $F$ and $C$. …

The question asks for the minimum distance between an object and its real image formed by a concave mirror. This is a classic problem that tests your understanding of the mirror formula and the concept of real images.

The core idea

A real image is formed when rays actually converge after reflection. For a concave mirror, a real image is formed only when the object is placed beyond the focus (i.e., $u > f$). The image distance $v$ is then positive (real) and given by the mirror formula:

The distance between the object and its image is $|u - v|$. We want to find the smallest possible value of this distance for real images.

Step-by-step reasoning

1. Set up the mirror formula For a concave mirror, $f$ is positive. Let $u$ be the object distance (positive, measured from the mirror). For a real image, $v$ is also positive. The mirror formula gives:

So:

2. Write the distance between object and image Let $D = |u - v|$. Since both $u$ and $v$ are positive and measured from the mirror, the object and image lie on the same side of the mirror. The distance between them is:

3. Simplify the expression Factor $u$:

Since for real images $u > f$, the denominator $u - f > 0$. The sign of $u - 2f$ depends on $u$. So:

4. Analyse the behaviour

   • If $u > 2f$, then $u - 2f > 0$, so $D = u \cdot \frac{u - 2f}{u - f}$.
   • If $f < u < 2f$, then $u - 2f < 0$, so $D = u \cdot \frac{2f - u}{u - f}$.

   In both cases, $D$ is positive. The question is: can $D$ become zero?

5. When does $D = 0$?

   $D = 0$ when $|u - 2f| = 0$, i.e., when $u = 2f$. …