Light Class 10 Important Numericals with Step by Step Solutions

Light is one of the most important and most scoring chapters in Class 10 Physics because it combines concepts, ray diagrams, formulas, sign conventions, and numericals in a very structured way. At the same time, it is also one of the chapters students fear the most because they have to remember image formation cases, mirror and lens formulas, magnification, refractive index, and the sign conventions for different situations.

The good news is that Light becomes much easier once students learn the summary tables properly and know how to solve numericals step by step. In most cases, students do not lose marks because the chapter is too difficult. They lose marks because they forget the sign of u, v, and f, or they panic and choose the wrong formula.

At Deeksha Vedantu, we always encourage students to solve Light numericals by following a fixed method: identify the mirror or lens, decide the sign convention, write the known values clearly, apply the correct formula, and then interpret the answer using the image-formation rules. This method makes the chapter far more manageable.

Why Light Numericals Are Important in Class 10

Light numericals are important because they appear regularly in CBSE board papers and they test both conceptual clarity and calculation accuracy.

Why Students Should Practise This Topic Well

• it is a regular board-exam chapter
• it includes numericals from mirrors, lenses, and refractive index
• it improves sign convention accuracy
• it helps students link theory with image formation
• it supports case-based and competency-style questions

What Students Must Revise Before Solving Numericals

Before solving any numerical from Light, students should revise the concept summary properly.

Four Most Important Areas to Remember

Students should remember the image formation rules for:

• concave mirror
• convex mirror
• convex lens
• concave lens

If these are clear, numericals become much easier.

Image Formation Summary Tables

These tables are useful for quick revision before solving numericals.

Concave Mirror Image Cases

Convex Mirror Image Cases

Convex Lens Image Cases

Concave Lens Image Cases

Sign Convention for Light Numericals

Students must not mix mirror and lens signs.

Sign Convention Table

Important Formulas for Light Numericals

Students should revise these formulas before starting practice.

Formula Summary Table

Meaning of Symbols in Refractive Index

Step-by-Step Method to Solve Any Numerical from Light

This is the safest exam method and works well in board exams because it reduces sign mistakes and wrong conclusions.

Step-by-Step Solving Table

Important Numerical 1: Convex Mirror Rear-View Mirror Question

A convex mirror used as a rear-view mirror has a radius of curvature of 3 m. If a bus is located 5 m from the mirror, find the position, nature, and size of the image.

Step 1: Identify the Mirror and Known Values

Since it is a convex mirror:

• R = +3 m
• u = -5 m

Now use:

f = R/2 = 3/2 m

So:

f = +1.5 m

Step 2: Use Mirror Formula

1/f = 1/v + 1/u

1/1.5 = 1/v + 1/(-5)

2/3 = 1/v – 1/5

1/v = 2/3 + 1/5

1/v = 10/15 + 3/15 = 13/15

v = 15/13 m

Step 3: Interpret the Result

Since v is positive, the image is formed behind the mirror.

So the image is:

• virtual
• erect
• diminished

Step 4: Find Magnification

m = -v/u

m = -(15/13)/(-5)

m = 3/13

Answer

• image position = 15/13 m behind the mirror
• nature = virtual and erect
• size = diminished because magnification is less than 1

Important Numerical 2: Concave Mirror and Screen Question

An object of height 4 cm is placed 25 cm in front of a concave mirror of focal length 15 cm. Find the position, nature, and size of the image.

Step 1: Write Known Values

For a concave mirror:

• u = -25 cm
• f = -15 cm
• hₒ = 4 cm

Step 2: Use Mirror Formula

1/f = 1/v + 1/u

1/(-15) = 1/v + 1/(-25)

-1/15 = 1/v – 1/25

1/v = -1/15 + 1/25

1/v = (-5 + 3)/75 = -2/75

v = -37.5 cm

Step 3: Interpret the Position

Since v is negative, the image is formed in front of the mirror.

So the image is:

• real
• inverted

Step 4: Find Magnification

m = -v/u

m = -(-37.5)/(-25) = -1.5

Now use:

m = hᵢ/hₒ

-1.5 = hᵢ/4

hᵢ = -6 cm

Answer

• image position = -37.5 cm
• nature = real and inverted
• image height = -6 cm

The negative image height confirms inversion.

Important Numerical 3: Magnification of Concave Mirror

A concave mirror produces a 3 times magnified real image of an object placed 10 cm in front of it. Where is the image located?

Step 1: Write Known Values

For a real image, magnification is negative.

So:

• m = -3
• u = -10 cm

Step 2: Use Magnification Formula

m = -v/u

-3 = -v/(-10)

-3 = v/10

v = -30 cm

Answer

The image is formed at 30 cm in front of the mirror.

Since v is negative, the image is real and inverted.

Important Numerical 4: Refractive Index Question

Light enters from air into glass of refractive index 1.5. Find the speed of light in the glass if the speed of light in air is 3 × 10⁸ m/s.

Step 1: Use Refractive Index Formula

n = c/v

Where:

• n = 1.5
• c = 3 × 10⁸ m/s
• v = speed in glass

Step 2: Substitute Values

1.5 = (3 × 10⁸)/v

v = (3 × 10⁸)/1.5

v = 2 × 10⁸ m/s

Answer

The speed of light in glass is 2 × 10⁸ m/s.

Important Numerical 5: Erect Image Using Concave Mirror

A student wants to obtain an erect image using a concave mirror of focal length 10 cm. At what distance should the object be placed from the mirror?

Step 1: Use the Image Formation Rule

A concave mirror forms an erect image only when the object is placed between the pole P and focus F.

Step 2: Interpret the Range

If focal length is 10 cm, then the object must be placed at a distance less than 10 cm from the mirror.

Answer

The object should be placed between the pole and focus, that is, at a distance less than 10 cm from the mirror.

Important Numerical 6: Concave Lens Problem

A concave lens has focal length 15 cm. At what distance should an object be placed so that the image is formed 10 cm from the lens? Also find the magnification.

Step 1: Write Known Values

For a concave lens:

• f = -15 cm
• v = -10 cm

We need to find u.

Step 2: Use Lens Formula

1/f = 1/v – 1/u

1/(-15) = 1/(-10) – 1/u

-1/15 = -1/10 – 1/u

-1/u = -1/15 + 1/10

-1/u = (-2 + 3)/30 = 1/30

u = -30 cm

Step 3: Find Magnification

m = v/u

m = (-10)/(-30)

m = 1/3

Answer

• object distance = -30 cm
• magnification = 1/3

Since magnification is positive and less than 1, the image is virtual, erect, and diminished.

Important Numerical 7: Identify the Mirror from Image Nature

A spherical mirror always forms an erect and diminished image for any object placed in front of it. Identify the mirror and draw the conclusion about the image.

Step 1: Use the Rule

Only a convex mirror always forms an erect and diminished image for all positions of the object.

Answer

The mirror is a convex mirror.

The image formed is always:

• virtual
• erect
• diminished

Important Numerical 8: Convex Lens and Screen Problem

A student focuses the image of a candle flame on a screen using a convex lens. The candle is at 12 cm, the lens is at 50 cm, and the screen is at 88 cm. Find the focal length of the lens.

Step 1: Find Object Distance from Lens

Lens position = 50 cm

Object position = 12 cm

So:

u = 12 – 50 = -38 cm

Step 2: Find Image Distance from Lens

Screen position = 88 cm

So:

v = 88 – 50 = +38 cm

Step 3: Use Lens Formula

1/f = 1/v – 1/u

1/f = 1/38 – 1/(-38)

1/f = 1/38 + 1/38

1/f = 2/38 = 1/19

f = 19 cm

Answer

The focal length of the convex lens is 19 cm.

Important Numerical 9: Shift the Object to the Focus of a Convex Lens

For the same convex lens of focal length 19 cm, where will the image be formed if the candle is moved to 31 cm from the same reference system?

Step 1: Understand the New Position

Lens is still at 50 cm.

New object position = 31 cm

So:

u = 31 – 50 = -19 cm

This means the object is at the focus.

Step 2: Use the Rule

For a convex lens, when the object is at the focus, the image is formed at infinity.

Answer

The image will be formed at infinity.

Important Numerical 10: Mirror with Magnification -1

A spherical mirror produces an image of magnification -1 on a screen placed at a distance of 40 cm. Identify the mirror, state the nature of image, and find the object position.

Step 1: Interpret the Given Information

If the image is formed on a screen, it must be a real image.

So the mirror must be a concave mirror.

Step 2: Use Magnification

For a real image in a mirror:

m = -v/u

Given:

m = -1

And the image is formed on screen at 40 cm, so:

v = -40 cm

Now:

-1 = -(-40)/u

-1 = 40/u

u = -40 cm

Step 3: Interpret the Result

Object and image are at the same distance.

This happens when the object is at the center of curvature C.

Answer

• mirror = concave mirror
• image nature = real and inverted
• object position = -40 cm

How to Find the Nature of Image from the Numerical Answer

Students often stop after finding v, but that is not enough. They must also identify the image nature.

Nature of Image from v Table

How to Find Image Nature from Magnification

This is another very useful shortcut.

Image Nature from Magnification Table

Common Mistakes Students Make in Light Numericals

Common Mistakes Table

Quick Revision Summary for Light Numericals

Must-Remember Formulas Table

Must-Remember Concept Rules Table

Best Strategy to Score Well in Light Numericals

Step 1: Learn the Four Image Summary Tables Properly

Without this, interpretation becomes hard.

Step 2: Memorise Sign Conventions

These are essential for correct calculation.

Step 3: Practise One Chapter-Wise Formula Sheet

Keep all mirror, lens, and refractive index formulas in one place.

Step 4: Solve NCERT Numericals First

These create the base for board-style questions.

Step 5: Move to PYQs and Competency Questions

This builds exam confidence.

Practice Questions for Students

Important Practice Questions

• A concave mirror has focal length 20 cm. Find the image position when the object is at 30 cm.
• A convex mirror has radius of curvature 4 m. Find its focal length.
• A convex lens forms an image at infinity. Where is the object placed?
• A concave lens has focal length 12 cm and object distance 24 cm. Find image position.
• The refractive index of a medium is 1.33. Find the speed of light in that medium.

FAQs

Q1. What is the most important thing in Light numericals?

The most important thing is the correct sign convention along with the proper choice of formula.

Q2. What is the mirror formula in Class 10?

The mirror formula is 1/f = 1/v + 1/u.

Q3. What is the lens formula in Class 10?

The lens formula is 1/f = 1/v – 1/u.

Q4. How do I know whether the image is real or virtual from the answer?

For mirrors, negative v means real image and positive v means virtual image. For lenses, positive v means real image and negative v means virtual image.

Q5. How do I know whether the image is erect or inverted?

A positive magnification means erect image, and a negative magnification means inverted image.

Q6. Why do students lose marks in Light numericals?

Students mostly lose marks because of sign convention errors, wrong formula choice, or incomplete final interpretation.

Q7. What is the formula for refractive index?

The formula is n = c/v.

Q8. How can I improve in Light numericals quickly?

Revise the four image-formation tables, memorise sign conventions, practise NCERT numericals first, and then solve previous year questions step by step.

Conclusion

Light is one of the most important Class 10 Physics chapters because it combines concept understanding with calculation-based problem solving. Once students know the image formation rules, sign conventions, and standard formulas, numericals become much easier to solve. The real key is to stay calm, identify the optical device correctly, and follow a proper step-by-step method.

The best way to prepare this chapter is to revise the summary tables regularly, solve numericals by hand, and always interpret the final answer properly. At Deeksha Vedantu, we always remind students that Light numericals are scoring once the basics are clear and the solving method is systematic.