Mastering Numerical Problems in Light: Mirrors and Lenses Class 10

In CBSE Class 10 Science, the physics section carries approximately $25$ marks, with Optics (Chapter 9: Light - Reflection and Refraction) contributing between $8$ and $10$ marks. Almost half of these marks come directly from numerical calculations involving spherical mirrors, lenses, magnification, and lens power.

Despite having only two core formulas—the Mirror Formula and the Lens Formula—students frequently forfeit marks due to a single recurrent error: assigning incorrect Cartesian signs ($+$ or $-$) to variables.

In this guide, we provide the foolproof sign assignment algorithm, decode the diagnostic meaning of magnification ($m$), and work through full-length solved board exam numericals.

What You Will Learn

• The Foolproof 4-Step Cartesian Sign Assignment Protocol
• The Golden Sign Rules: $u, v, f$ for Concave and Convex systems
• Mirror Formula vs. Lens Formula: Avoiding the sign trap
• Magnification Decoding: How sign and magnitude reveal the nature and size of an image
• Solving "Screen Distance" problems
• Solving "Magnifying Glass / Virtual Image" problems
• Power of a Lens and combinations of thin lenses

1. The Foolproof Cartesian Sign Protocol

Before substituting a single number into your formula, write down your given data table and apply these non-negotiable sign rules:

                            The Golden Cartesian Rules
    -------------------------------------------------------------------------
    Variable            Concave Mirror / Lens       Convex Mirror / Lens
    -------------------------------------------------------------------------
    Object Distance (u) ALWAYS NEGATIVE (-)         ALWAYS NEGATIVE (-)
    Focal Length (f)    ALWAYS NEGATIVE (-)         ALWAYS POSITIVE (+)
    Image Distance (v)  Negative (Real image)       Always Positive (+) [Mirror]
                        Positive (Virtual image)    Positive (Real image) [Lens]
    Image Height (h')   Negative (Inverted)         Positive (Erect)
    -------------------------------------------------------------------------

The Universal Rule for Object Distance: <u>Object distance ($u$) is ALWAYS NEGATIVE ($-$) in EVERY single numerical problem for all mirrors and all lenses, because the object is always placed to the left of the optical surface!</u>

2. Mirrors vs. Lenses: The Formula Contrast

    Optical System        Fundamental Formula                  Linear Magnification
    -------------------------------------------------------------------------------
    Spherical MIRROR      1/v + 1/u = 1/f   (PLUS sign)        m = -v/u   (MINUS sign)
    Spherical LENS        1/v - 1/u = 1/f   (MINUS sign)       m = +v/u   (PLUS sign)

3. Magnification Diagnostic Decoder ($m$)

The linear magnification value $m$ tells you the complete optical story:

1. The Sign of $m$ reveals the NATURE of the image:
   • $m < 0$ (Negative): The image is Real and Inverted.
   • $m > 0$ (Positive): The image is Virtual and Erect.
2. The Absolute Magnitude of $|m|$ reveals the SIZE of the image:
   • $|m| > 1$: The image is enlarged (magnified).
   • $|m| = 1$: The image is the same size as the object (object at $C$ or $2F$).
   • $|m| < 1$: The image is diminished.

4. Solved CBSE Board Examination Problems

Solved Problem 1: Concave Mirror "Screen Distance" Numerical

Problem: A $2.0\text{ cm}$ tall object is placed at a distance of $30\text{ cm}$ from a concave mirror of focal length $20\text{ cm}$. At what distance from the mirror should a screen be placed to obtain a sharp image? Find the nature and height of the image.

Solution:

1. Assign Cartesian Signs:
   • Height of object: $h = +2.0\text{ cm}$.
   • Object distance: $u = \mathbf{-30\text{ cm}}$ (Always negative).
   • Focal length of concave mirror: $f = \mathbf{-20\text{ cm}}$ (Concave $\implies$ negative).
   • Image distance: $v = ?$ and Image height: $h' = ?$
2. Apply the Mirror Formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \implies \frac{1}{v} = \frac{1}{f} - \frac{1}{u}$ $\frac{1}{v} = \frac{1}{-20} - \left(\frac{1}{-30}\right) = -\frac{1}{20} + \frac{1}{30}$ LCM of 20 and 30 is 60: $\frac{1}{v} = \frac{-3 + 2}{60} = \frac{-1}{60} \implies \mathbf{v = -60\text{ cm}}$
3. Calculate Magnification and Image Height: $m = \frac{h'}{h} = -\frac{v}{u}$ $h' = -\frac{v \times h}{u} = -\frac{(-60) \times (+2.0)}{-30} = -\frac{-120}{-30} = \mathbf{-4.0\text{ cm}}$
4. Conclusion:
   • The screen must be placed $60\text{ cm}$ in front of the mirror.
   • Since $v < 0$ and $h' < 0$, the image is Real and Inverted.
   • Since $|h'| = 4.0\text{ cm} > 2.0\text{ cm}$, the image is enlarged to twice the object's height.

Solved Problem 2: Convex Lens Magnifying Glass (Virtual Image)

Problem: A convex lens of focal length $10\text{ cm}$ forms a virtual image $25\text{ cm}$ away from the lens. Find how far the object is placed from the lens, and calculate its magnification.

Solution:

1. Assign Cartesian Signs:
   • Focal length of convex lens: $f = \mathbf{+10\text{ cm}}$ (Convex $\implies$ positive).
   • Image is virtual $\implies$ formed on the same side as the object $\implies v = \mathbf{-25\text{ cm}}$.
   • Object distance $u = ?$ and Magnification $m = ?$
2. Apply the Lens Formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \implies \frac{1}{u} = \frac{1}{v} - \frac{1}{f}$ $\frac{1}{u} = \frac{1}{-25} - \frac{1}{10} = -\frac{1}{25} - \frac{1}{10}$ LCM of 25 and 10 is 50: $\frac{1}{u} = \frac{-2 - 5}{50} = \frac{-7}{50} \implies u = -\frac{50}{7} = \mathbf{-7.14\text{ cm}}$
3. Calculate Magnification ($m$): $m = +\frac{v}{u} = \frac{-25}{-50/7} = \frac{-25 \times 7}{-50} = \mathbf{+3.5}$
4. Conclusion:
   • The object is placed at a distance of $7.14\text{ cm}$ in front of the lens.
   • Since $m = +3.5$ is positive, the image is Virtual and Erect.
   • Since $|m| = 3.5 > 1$, the image is magnified $3.5$ times.

5. Summary and Examination Tips

Exam Tip: Whenever a problem states: "An image is caught on a screen", it is telling you that the image is REAL ($v < 0$ for mirrors, $v > 0$ for lenses)! Virtual images cannot be caught on a screen.

Common Mistake: Forgetting to write units in your final answer. Writing $v = -60$ without $\text{cm}$ costs half a mark on board papers!