Q.Use the mirror equation to deduce that:

Mirror equation $\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$; magnification $m = -\dfrac{v}{u}$. Convention: concave $f<0$, convex $f>0$, real object $u<0$; $v<0$ is a real image (in front), $v>0$ a virtual image (behind).

(a) Concave, object between $f$ and $2f$ ($2f < u < f$, all $<0$): $\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u}$. Reciprocating $2f < u < f$ gives $\dfrac{1}{f} < \dfrac{1}{u} < \dfrac{1}{2f}$, so $\dfrac{1}{2f} < \dfrac{1}{v} < 0$: thus $v<0$ (real) and $|v| > 2|f|$ — real image beyond $2f$.

(b) Convex ($f>0$, $u<0$): $\dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{|u|} > 0$, so $v>0$ always — image always virtual.

(c) Convex: $v = \dfrac{f|u|}{f+|u|} < f$, so $0<v<f$ (between pole and focus). $m = \dfrac{v}{|u|} = \dfrac{f}{f+|u|} < 1$ — diminished. …

Convention used: distances measured from the pole; real object $u<0$; concave mirror $f<0$, convex mirror $f>0$; a real image has $v<0$ (formed in front of the mirror), a virtual image has $v>0$ (behind). Magnification $m=-\dfrac{v}{u}$; $|m|>1$ enlarged, $|m|<1$ diminished.

(a) Concave mirror, object between $f$ and $2f$

Here $f<0$ and the object lies between $f$ and $2f$: $2f < u < f$ (all negative). From the mirror equation,

Taking reciprocals of $2f < u < f$ (order reverses for negatives): $\dfrac{1}{f} < \dfrac{1}{u} < \dfrac{1}{2f}$. Subtracting from $\dfrac{1}{f}$,

So $\dfrac{1}{v}<0 \Rightarrow v<0$: the image is real. And $\dfrac{1}{v} > \dfrac{1}{2f}$ with both negative means $|v| > 2|f|$: the image lies beyond $2f$.

(b) Convex mirror — always virtual

Now $f>0$ and $u<0$, so

for every object position. A positive $v$ means the image is behind the mirror — virtual, independent of where the object is.

(c) Convex mirror — diminished, between pole and focus …

Method: Algebraic Deduction Using the Mirror Equation

Method Name: Sign Convention–Based Algebraic Analysis

Steps:

1. Write the mirror equation:

2. Apply the Cartesian sign convention:
   • For concave mirror: $f$ is negative ($f < 0$)
   • For convex mirror: $f$ is positive ($f > 0$)
   • Object distance $u$ is always negative ($u < 0$)
   • Image distance $v$: positive means real image, negative means virtual image
3. Solve for $v$ in terms of $u$ and $f$:

4. Use the sign of $v$ to determine real/virtual and magnification $m = -\frac{v}{u}$ to determine size and orientation.

(a) Concave mirror: object between $f$ and $2f$ → real image beyond $2f$

• Given: $f < 0$, $u$ is negative, and $|f| < |u| < 2|f|$
• From mirror equation:

Since $|u| > |f|$, $\frac{1}{|u|} < \frac{1}{|f|}$, so $\frac{1}{v}$ is negative → $v$ is negative → real image

• Magnitude: $|v| > 2|f|$ because $\frac{1}{|v|} = \frac{1}{|f|} - \frac{1}{|u|}$ and $|u| < 2|f|$ gives $\frac{1}{|v|} < \frac{1}{2|f|}$ → $|v| > 2|f|$

Result: Real image beyond $2f$.

(b) Convex mirror always produces a virtual image

• Given: $f > 0$, $u < 0$
• Mirror equation:

So $v > 0$ → virtual image (since $v$ positive means behind the mirror)

Result: Virtual image for any object position.

(c) Convex mirror: image diminished, between pole and focus

• From (b), $v > 0$ and $\frac{1}{v} = \frac{1}{f} + \frac{1}{|u|}$
• Since $\frac{1}{v} > \frac{1}{f}$, we get $v < f$ → image lies between pole and focus
• Magnification: …

Common Mistakes with the Spherical Mirror Equation

Students often struggle with algebraic deductions from the mirror formula. Here are the most frequent errors and how to avoid them.

Mistake 1: Forgetting the Sign Convention

The Error:

Using $u$ (object distance) as positive for concave mirrors or treating $f$ as positive for convex mirrors.

Why It Happens:

Students memorise the formula $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$ but ignore the Cartesian sign convention:

• Concave mirror: $f$ is negative ($f = -|f|$)
• Convex mirror: $f$ is positive ($f = +|f|$)
• Object distance $u$ is always negative (object in front of mirror)

How to Avoid:

Always write the sign explicitly before substituting. For a concave mirror:

Mistake 2: Confusing the Range of $u$ for Concave Mirrors

The Error:

For part (a), students substitute $u = -f$ or $u = -2f$ directly instead of using inequalities.

Why It Happens:

The problem asks to deduce that an object between $f$ and $2f$ produces an image beyond $2f$. Plugging exact values gives only boundary cases.

How to Avoid:

Use the inequality:

Then solve for $v$ using:

and show that $|v| > 2|f|$ and $v$ is negative (real image).

Mistake 3: Assuming $v$ is Always Negative for Concave Mirrors

The Error:

Thinking all images from concave mirrors are real ($v$ negative).

Why It Happens:

Ray diagrams for concave mirrors show real images for objects beyond focus, but virtual images occur when the object is between pole and focus.

How to Avoid:

Check the sign of $v$ from the formula:

• If $v$ is negative → real image (in front of mirror)
• If $v$ is positive → virtual image (behind mirror)

For part (d): object between pole and focus means $|u| < |f|$. Substituting gives $v > 0$ → virtual and enlarged.

Mistake 4: Misinterpreting "Diminished" for Convex Mirrors

The Error:

Stating that the image is diminished only for certain object positions.

Why It Happens:

Students recall that convex mirrors always give diminished images but fail to prove it algebraically.

How to Avoid:

Use magnification:

For convex mirrors, $f > 0$, $u < 0$, and $v > 0$ (always virtual). Show that:

because $v < |u|$ always holds for convex mirrors.

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