Q.Show that if an object is placed at a distance 'x' in front of a concave mirror of radius of curvature 'r', its image is formed at a distance rx/(2x–r) in front of the mirror. OR In Young's double-slit experiment, when light of wavelength 600 nm is used, 12 fringes are formed within a certain portion of the screen. If light of wavelength 400 nm is used instead, how many fringes will be formed within that same portion of the screen?

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 $u = 2f$. …