CBSE Class 12 Physics 2026: question 32(b)

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1. (a) Ray diagram (described in words). Draw a concave mirror with pole $P$, focus $F$ and centre of curvature $C$ on the principal axis. Place a small object $AB$ with $B$ on the axis between $F$ and $C$ and $A$ above the axis.

   • Ray 1: from $A$ parallel to the axis, striking the mirror at $M$ and reflecting through $F$.
   • Ray 2: from $A$ to the pole $P$, reflecting so that it makes an equal angle with the axis on the other side.

   The reflected rays meet at $A^{\prime }$ below the axis, beyond $C$. Draw the perpendicular $A^{\prime }{B}^{\prime }$ to the axis. The image $A^{\prime }{B}^{\prime }$ is real, inverted and magnified.

   1 mark

2. Similar triangles. Consider paraxial rays, so $M$ is close to $P$ and $MP$ can be taken as a straight line perpendicular to the axis, with $MP=AB$.

   • $△A^{\prime }{B}^{\prime }F\sim △MPF$ (equal vertically opposite angles at $F$, and right angles at $B^{\prime }$ and $P$). Hence $\frac{B^{\prime }{A}^{\prime }}{PM}=\frac{B^{\prime }F}{FP}$, which gives $\frac{B^{\prime }{A}^{\prime }}{BA}=\frac{B^{\prime }F}{FP}$ ...(1)
   • $△A^{\prime }{B}^{\prime }P\sim △ABP$ (angle of incidence equals angle of reflection at $P$, and right angles at $B$ and $B^{\prime }$). Hence $\frac{B^{\prime }{A}^{\prime }}{BA}=\frac{B^{\prime }P}{BP}$ ...(2)

   From (1) and (2):
   $$\frac{B^{\prime }F}{FP}=\frac{B^{\prime }P}{BP},\text{where }{B}^{\prime }F=B^{\prime }P-FP.$$

   1 mark

3. Apply the sign convention. All distances are measured from $P$, and distances against the incident light are negative. So $BP=-u$, $B^{\prime }P=-v$ and $FP=-f$.

   Substituting:
   $$\frac{-v+f}{-f}=\frac{-v}{-u}$$
   $$\Rightarrow \frac{v-f}{f}=\frac{v}{u}$$
   $$\Rightarrow \frac{v}{f}-1=\frac{v}{u}$$

   Dividing throughout by $v$:
   $$\frac{1}{f}-\frac{1}{v}=\frac{1}{u}$$
   $$\Rightarrow \boxed{\frac{1}{v}+\frac{1}{u}=\frac{1}{f}}$$

   This is the mirror formula.

   1 mark

4. (b) Find the image distance. The image is virtual and erect, so $m=+2$. The object is in front of the mirror, so $u=-10$ cm.

   $$m=-\frac{v}{u}=2$$
   $$\Rightarrow v=-2u=-2(-10)=+20\text{ cm}$$

   The image is 20 cm behind the mirror.

   1 mark

5. Find the focal length.
   $$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}=\frac{1}{20}+\frac{1}{-10}=\frac{1-2}{20}=-\frac{1}{20}$$
   $$\Rightarrow f=-20\text{ cm}$$

   The focal length of the concave mirror is 20 cm.

   1 mark

Answer: (a) Mirror formula $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$; (b) $f=-20$ cm (focal length 20 cm)

Revise the chapter

• Chapter: Ray Optics and Optical Instruments (notes, formulas, where marks are lost and every past question)