Mirror Formula, Sign Convention, and Linear Magnification for CBSE Class 10

In ray optics, ray diagrams provide a beautiful visual representation of how images are formed. However, when an optical engineer designs a telescope, an optometrist fits corrective optics, or a student solves board examination questions, visual drawings must be backed by exact mathematical calculations. Where will the image form? How tall will it be? Will it be real or virtual?

In CBSE Class 10 Science, Chapter 9 (Light - Reflection and Refraction), these quantitative questions are resolved by three mathematical tools: the New Cartesian Sign Convention, the Mirror Formula, and the Linear Magnification Equation.

What You Will Learn

• The New Cartesian Sign Convention rules for spherical mirrors
• The golden focal length rules ($f$ is negative for concave, positive for convex)
• The Mirror Formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$
• The Linear Magnification Formula: $m = \frac{h'}{h} = -\frac{v}{u}$
• How to decode the sign and numerical magnitude of magnification ($m$)
• Step-by-step solved CBSE board exam numerical problems
• Common algebraic and sign pitfalls

1. The New Cartesian Sign Convention

To calculate distances algebraically, we adopt the New Cartesian Sign Convention, where the mirror's Pole ($P$) is placed at the origin $(0, 0)$ of a coordinate system:

                            Height Upwards (+y)
                                    ^
                    Direction of    |
                    Incident Light  |
                   ---------------->| (Pole P = Origin)
                    Distance Left   |   Distance Right
                      (-x axis)     |     (+x axis)
                   <----------------+---------------->
                                    |
                                    v
                           Height Downwards (-y)

The Five Sign Rules:

1. Object Placement: The object is always placed to the left of the mirror, meaning light always travels from left to right.
2. Origin: All distances are measured strictly from the Pole ($P$) of the mirror as origin.
3. Horizontal Distances:
   • Distances measured in the direction of incident light (along $+x$-axis, to the right of $P$) are positive ($+$).
   • Distances measured against the direction of incident light (along $-x$-axis, to the left of $P$) are negative ($-$).
4. Vertical Heights:
   • Heights measured upwards perpendicular to the principal axis (along $+y$-axis) are positive ($+$).
   • Heights measured downwards perpendicular to the principal axis (along $-y$-axis) are negative ($-$).

The Golden Sign Rules (Must Memorize!):

1. Object Distance ($u$): Always NEGATIVE ($-$) for all mirrors.
2. Focal Length of Concave Mirror ($f$): Always NEGATIVE ($-$) ($F$ lies in front).
3. Focal Length of Convex Mirror ($f$): Always POSITIVE ($+$) ($F$ lies behind).
4. Image Distance ($v$):
   • If $v$ is negative ($-$) $\implies$ Image is in front of mirror $\implies$ Real and Inverted.
   • If $v$ is positive ($+$) $\implies$ Image is behind mirror $\implies$ Virtual and Erect.

2. The Mirror Formula

The algebraic relationship connecting the object distance ($u$), image distance ($v$), and focal length ($f$) of a spherical mirror is called the Mirror Formula:

The Mirror Formula

$\mathbf{\frac{1}{v} + \frac{1}{u} = \frac{1}{f}}$

Where:

• $u = \text{Object distance from Pole } P$
• $v = \text{Image distance from Pole } P$
• $f = \text{Focal length of the mirror } (f = R/2)$

Important: <u>While substituting numerical values into the mirror formula, you MUST attach the appropriate positive or negative signs to each known variable according to the sign convention!</u>

3. Linear Magnification ($m$)

The ratio of the height of the image to the height of the object is called linear magnification:

Magnification Formula

$\mathbf{m = \frac{\text{Height of Image } (h')}{\text{Height of Object } (h)} = -\frac{v}{u}}$

How to Decode the Magnification Value ($m$):

A. The Sign of $m$:

• If $m$ is Negative ($-$): The image is Real and Inverted ($h'$ is negative).
• If $m$ is Positive ($+$): The image is Virtual and Erect ($h'$ is positive).

B. The Numerical Magnitude of $|m|$:

• If $|m| > 1$: Image is magnified (enlarged) (larger than object).
• If $|m| = 1$: Image is the same size as the object (object at $C$).
• If $|m| < 1$: Image is diminished (smaller than object).

4. Solved CBSE Board Examination Problems

Solved Example 1: Convex Rear-View Mirror (NCERT Classic)

Problem: A convex mirror used for rear-view on an automobile has a radius of curvature of $3.00\text{ m}$. If a bus is located at $5.00\text{ m}$ from this mirror, find the position, nature, and size of the image.

Solution:

1. List Given Data with Cartesian Signs:
   • Radius of curvature $R = +3.00\text{ m}$ (Convex mirror).
   • Focal length $f = \frac{R}{2} = \frac{+3.00}{2} = \mathbf{+1.50\text{ m}}$.
   • Object distance $u = \mathbf{-5.00\text{ m}}$.
   • Image distance $v = ?$ and Magnification $m = ?$
2. Apply the Mirror Formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \implies \frac{1}{v} = \frac{1}{f} - \frac{1}{u}$ $\frac{1}{v} = \frac{1}{+1.50} - \left(\frac{1}{-5.00}\right) = \frac{1}{1.5} + \frac{1}{5.0} = \frac{10}{15} + \frac{1}{5} = \frac{2}{3} + \frac{1}{5} = \frac{10 + 3}{15} = \frac{13}{15}$ $v = \frac{15}{13} = \mathbf{+1.15\text{ m}}$
3. Calculate Magnification ($m$): $m = -\frac{v}{u} = -\frac{+1.15}{-5.00} = \mathbf{+0.23}$
4. Interpret the Results:
   • Since $v = +1.15\text{ m}$ (positive), the image is formed $1.15\text{ m}$ behind the mirror.
   • Since $m$ is positive ($+$), the image is Virtual and Erect.
   • Since $|m| = 0.23 < 1$, the image is diminished to roughly $23\%$ of the bus's actual size.

Solved Example 2: Concave Mirror Real Image Problem

Problem: An object $4.0\text{ cm}$ in size is placed at $25.0\text{ cm}$ in front of a concave mirror of focal length $15.0\text{ cm}$. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Find the nature and the size of the image.

Solution:

1. List Given Data with Signs:
   • Object height $h = +4.0\text{ cm}$.
   • Object distance $u = -25.0\text{ cm}$.
   • Focal length of concave mirror $f = -15.0\text{ cm}$.
   • Screen distance $v = ?$ and Image height $h' = ?$
2. Apply the Mirror Formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \implies \frac{1}{v} = \frac{1}{f} - \frac{1}{u}$ $\frac{1}{v} = \frac{1}{-15} - \left(\frac{1}{-25}\right) = -\frac{1}{15} + \frac{1}{25}$ Taking LCM of 15 and 25 (which is 75): $\frac{1}{v} = \frac{-5 + 3}{75} = \frac{-2}{75}$ $v = -\frac{75}{2} = \mathbf{-37.5\text{ cm}}$
3. Calculate Image Height ($h'$): $m = \frac{h'}{h} = -\frac{v}{u} \implies h' = -\frac{v \times h}{u}$ $h' = -\frac{(-37.5) \times (+4.0)}{-25.0} = -\frac{150}{-25} = \mathbf{-6.0\text{ cm}}$
4. Conclusion:
   • The screen must be placed $37.5\text{ cm}$ in front of the mirror.
   • The image is Real and Inverted (since $v < 0$ and $h' < 0$).
   • The image is enlarged ($6.0\text{ cm}$ tall).

5. Summary and Examination Tips

Exam Tip: In questions asking "Where should a screen be placed to capture the image?", they are asking you to find $v$! Screens can ONLY capture real images; virtual images cannot be projected onto a screen.

Common Mistake: Forgetting the minus sign in the magnification formula ($m = -v/u$). In mirrors, the formula has a minus sign: $m = -v/u$. In lenses, it is positive: $m = +v/u$!