Q.An object is placed at

Using the spherical mirror equation $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ with $f=R/2=7.5\ \text{cm}$ (concave, so $f=-7.5\ \text{cm}$):

Case (i), $u=-10\ \text{cm}$: $\frac{1}{v}=\frac{1}{-7.5}-\frac{1}{-10}=-\frac{1}{30}$, so $v=-30\ \text{cm}$ (real, in front of mirror). $m=-\frac{v}{u}=-3$ (inverted, enlarged). …

Setting up — the spherical mirror equation

The governing relation for any spherical mirror is

with the New Cartesian sign convention: distances are measured from the mirror's pole, with the direction of incident light taken positive. A real object placed in front of the mirror always has $u$ negative, and a concave mirror has its focus in front of it, so $f$ is also negative.

Finding the focal length

Case (i): object at $10\ \text{cm}$

Here $u=-10\ \text{cm}$.

The negative sign means the image forms in front of the mirror — a real image.

Magnification:

The negative sign means the image is inverted, and $|m|=3$ means it is magnified 3 times.

This matches the expected behaviour: the object ($u=10\ \text{cm}$) lies between $F$ ($7.5\ \text{cm}$) and $C$ ($15\ \text{cm}$), so the image is real, inverted, and enlarged, beyond $C$ — exactly what we found.

Case (ii): object at $5\ \text{cm}$

Here $u=-5\ \text{cm}$.

The positive sign means the image forms behind the mirror — a virtual image.

Magnification: …

Method: Mirror Formula & Sign Convention (Cartesian)

This is the standard Cartesian sign convention method used in board exams. We'll apply the mirror formula and magnification formula step-by-step.

Step 1 — Write the given data with proper signs

For a concave mirror:

• Focal length $f$ is negative: $f = -\frac{R}{2}$
• Radius of curvature $R = 15\ \text{cm}$ $\Rightarrow f = -\frac{15}{2} = -7.5\ \text{cm}$

Object distance $u$ is always negative (object in front of mirror):

• Case (i): $u = -10\ \text{cm}$
• Case (ii): $u = -5\ \text{cm}$

Step 2 — Apply the mirror formula

Mirror formula:

Rearrange for $v$:

Case (i): $u = -10\ \text{cm}$, $f = -7.5\ \text{cm}$

Take LCM = 30:

So:

Nature: $v$ is negative → real image (in front of mirror).

Case (ii): $u = -5\ \text{cm}$, $f = -7.5\ \text{cm}$

Take LCM = 15:

So:

Nature: $v$ is positive → virtual image (behind the mirror).

Step 3 — Find magnification

Magnification formula:

Case (i): $v = -30$, $u = -10$

• $|m| > 1$ → enlarged
• $m$ negative → inverted image

Case (ii): $v = +15$, $u = -5$

• $|m| > 1$ → enlarged …

Common Mistakes & How to Avoid Them — Partial Mirror Obstruction

This problem is about standard mirror formula application, not partial obstruction. The "partial mirror" idea often confuses students into thinking only part of the mirror is used — but the mirror formula remains unchanged regardless of which portion of the mirror is exposed.

Here are the most frequent errors students make:

1. Confusing Radius of Curvature ($R$) with Focal Length ($f$)

Mistake: Using $R = 15\ \text{cm}$ directly in the mirror formula as $f$.

Why it happens: Students rush and don't recall the relation.

How to avoid: Always write:

Key point: For a concave mirror, $f$ is negative by sign convention:

2. Sign Convention Errors (Most Common)

Mistake: Using $u = +10\ \text{cm}$ or $u = +5\ \text{cm}$.

Why it happens: Students forget that object distance is always negative in the Cartesian sign convention (object is in front of the mirror).

How to avoid:

• Object is placed in front of mirror → $u$ is negative
• For case (i): $u = -10\ \text{cm}$
• For case (ii): $u = -5\ \text{cm}$

Correct formula:

3. Thinking "Partial Mirror" Changes the Formula

Mistake: Students believe that if only half the mirror is exposed, the image position changes.

Why it happens: Misunderstanding of "partial obstruction" concept.

How to avoid:

• Partial obstruction only reduces image brightness, not its position or nature
• The mirror formula works for any portion of the mirror
• Treat it as a full mirror for calculations

4. Incorrect Magnification Sign

Mistake: Writing $m = -\frac{v}{u}$ but forgetting to substitute signs correctly.

How to avoid:

• Always substitute $u$ and $v$ with their signs
• $m$ negative → real, inverted image
• $m$ positive → virtual, erect image

5. Not Checking Object Position Relative to Focus

Mistake: Applying the same logic for both cases without checking.

How to avoid: …