Light Class 10 Numericals Important Questions with Solutions for CBSE Boards

Light is one of the highest-weightage chapters in Class 10 Science, especially for numericals. In CBSE board exams, students are frequently asked questions based on mirrors, lenses, magnification, image nature, refractive index, and sign convention. That is why this chapter needs more than just theoretical reading. It requires careful formula use, proper sign handling, and repeated practice with numerical patterns.

Many students lose marks in Light not because the chapter is too difficult, but because they make small errors in sign convention, formula selection, or image interpretation. Once students understand how to identify the type of mirror or lens, how to use the mirror formula or lens formula, and how to interpret magnification, the chapter becomes much more scoring.

At Deeksha Vedantu, we always guide students to revise Light numericals in a structured way. When concepts are revised from basic to advanced patterns, students become more confident with board-style questions.

Why Light Numericals Are Important for CBSE Boards

Light is a chapter where both concept and calculation are tested together.

Why This Chapter Deserves Extra Focus

• It is a high-weightage chapter in Class 10 Physics.
• CBSE often asks direct and application-based numericals from it.
• Mirror and lens questions are repeated in similar patterns.
• Sign convention mistakes can easily change the final answer.
• Ray diagram understanding improves numerical accuracy.

What Students Must Know Before Solving Light Numericals

Before attempting important numericals from Light, students should revise the following basics.

Sign Convention

Sign convention is the most important foundation for this chapter.

For Mirrors

• Object distance u is generally negative.
• For concave mirror, focal length f is negative.
• For convex mirror, focal length f is positive.
• Real image distance is negative in mirror questions because it forms in front of the mirror.
• Virtual image distance is positive in mirror questions because it forms behind the mirror.

For Lenses

• Object distance u is generally negative.
• For convex lens, focal length f is positive.
• For concave lens, focal length f is negative.
• Real image distance is positive for lenses because it forms on the right side.
• Virtual image distance is negative for lenses because it forms on the left side.

Core Formulas from the Chapter Light

Students should memorise these formulas clearly.

Magnification and Height Relation

In the relation:

m = hᵢ/hₒ

Where:

• hᵢ = height of image
• hₒ = height of object

Quick Revision Logic for Image Nature

Students should not solve every question only by formula. Some answers can be identified directly through concept.

Important Numerical 1: Convex Mirror Used as Rear-View Mirror

A convex mirror used for rear view on an automobile has radius of curvature 3 m. A bus is located 5 m from this mirror. Find the position, nature, and size of the image.

Step 1: Write the Given Values

• Type of mirror = convex mirror
• Radius of curvature R = 3 m
• Object distance u = -5 m

Since f = R/2,

f = 3/2 = 1.5 m

For a convex mirror, focal length is positive.

So,

f = +1.5 m

Step 2: Apply Mirror Formula

1/f = 1/v + 1/u

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

1/v = 1/1.5 + 1/5

On solving,

v = 15/13 approximately 1.15 m

Step 3: Nature of Image

Since it is a convex mirror, the image is always virtual and erect.

Step 4: Magnification

m = -v/u

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

m is positive and less than 1, so the image is diminished.

Answer

• Position of image = approximately 1.15 m behind the mirror
• Nature = virtual and erect
• Size = diminished

Important Numerical 2: Concave Mirror with Object of Height 4 cm

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

Step 1: Given Values

• Type of mirror = concave mirror
• Object height ho = 4 cm
• Object distance u = -25 cm
• Focal length f = -15 cm

Step 2: Apply Mirror Formula

1/f = 1/v + 1/u

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

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

On solving,

v = -37.5 cm

Step 3: Find Magnification

m = -v/u

m = -(-37.5) divided by (-25)

m = -1.5

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

Step 4: Find Height of Image

m = hi/ho

-1.5 = hi/4

hi = -6 cm

Answer

• Position of image = -37.5 cm
• Nature = real and inverted
• Height of image = -6 cm

Important Numerical 3: Image Magnified Three Times by a Concave Mirror

A concave mirror produces three times magnified real image of an object placed at 10 cm. Where is the image located?

Given

• Magnification m = -3 because image is real
• Object distance u = -10 cm

Use Magnification Formula

m = -v/u

-3 = -v/(-10)

-3 = v/10

v = -30 cm

Answer

The image is located 30 cm in front of the mirror.

Important Numerical 4: Speed of Light in Glass

Light enters from air to glass having refractive index 1.5. The speed of light in vacuum is 3 × 10 raised to power 8 m/s. Find the speed of light in glass.

Given

• Refractive index of glass n = 1.5
• Speed of light in vacuum = 3 × 10 raised to power 8 m/s

Use Formula

n = speed of light in vacuum / speed of light in medium

1.5 = 3 × 10 raised to power 8 divided by v

v = 2 × 10 raised to power 8 m/s

Answer

The speed of light in glass is 2 × 10 raised to power 8 m/s.

Important Numerical 5: Erect Image Using Concave Mirror

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

Concept Used

A concave mirror forms an erect image only in one case.

That happens when the object is placed between:

• pole
• focus

Answer

The object should be placed between the pole and the focus of the concave mirror.

Important Numerical 6: Concave Lens with Given Image Distance

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

Step 1: Given Values

• Type of lens = concave lens
• Focal length f = -15 cm
• Image distance v = -10 cm because image formed by concave lens is virtual

Step 2: Apply Lens Formula

1/f = 1/v – 1/u

1/(-15) = 1/(-10) – 1/u

On solving,

u = -30 cm

Step 3: Magnification

m = v/u

m = (-10) divided by (-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

If a spherical mirror forms an image that is always erect and diminished for all positions of the object, identify the mirror.

Concept Used

Only a convex mirror always forms an image that is:

• virtual
• erect
• diminished

Answer

The mirror is a convex mirror.

Ray Diagram Logic

In a convex mirror, reflected rays diverge and appear to meet behind the mirror between the pole and the focus. That is why the image is always virtual, erect, and smaller.

Important Numerical 8: Refraction Through a Glass Slab

A student traces the path of light through a glass slab and measures angle of incidence, angle of refraction, and angle of emergence. What conclusion is correct?

Concept Used

When light enters glass from air:

• it bends towards the normal
• angle of refraction becomes smaller than angle of incidence

When it emerges back into air:

• it bends away from the normal
• angle of emergence becomes nearly equal to angle of incidence

Conclusion

Angle of incidence is greater than angle of refraction, but nearly equal to angle of emergence.

Correct Relation

• i is greater than r
• i is approximately equal to e

Important Numerical 9: Convex Lens with Given Candle, Lens, and Screen Positions

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

Step 1: Find Object Distance

Distance between candle and lens:

u = 50 – 12 = 38 cm

With sign convention,

u = -38 cm

Step 2: Find Image Distance

Distance between screen and lens:

v = 88 – 50 = 38 cm

For a convex lens forming real image,

v = +38 cm

Step 3: Interpret the Situation

Since object distance and image distance are equal, the object is placed at 2f and the image is also formed at 2f.

So,

2f = 38 cm

f = 19 cm

Answer

The focal length of the convex lens is 19 cm.

Important Numerical 10: Candle Shifted Toward the Convex Lens

In the same setup, the candle is shifted towards the lens to a position of 31 cm. Find where the image will be formed.

Step 1: New Object Distance

Lens is still at 50 cm.

Candle is now at 31 cm.

So,

u = 50 – 31 = 19 cm

With sign convention,

u = -19 cm

Step 2: Use Focal Length

From the previous part,

f = +19 cm

So here the object is placed at the focus of the convex lens.

Conclusion

When an object is placed at the focus of a convex lens, the image is formed at infinity.

If the Object Is Shifted Even Closer

If the object is moved further towards the lens and comes between the optical centre and focus, then the image becomes:

• virtual
• erect
• magnified

Important Numerical 11: Mirror with Magnification Minus One

A spherical mirror produces an image of magnification -1 on a screen placed 40 cm from the mirror. Identify the mirror, state the nature of image, and find how far the object is located from the mirror.

Step 1: Understand Magnification

m = -1 means:

• image is real
• image is inverted
• image is of the same size as the object

This happens in a concave mirror when the object is at the centre of curvature.

Step 2: Identify the Mirror

Since the image is real and can be formed on a screen, the mirror is concave.

Step 3: Find Object Position

Using

m = -v/u

-1 = -v/u

This gives:

v = u

Image distance is given as 40 cm in front of the mirror.

So,

v = -40 cm

Therefore,

u = -40 cm

Answer

• Mirror type = concave mirror
• Nature of image = real and inverted
• Object distance = -40 cm

Top Numerical Patterns Students Must Practise from Light

Light numericals often come in repeated question patterns.

Pattern 1: Find Position, Nature, and Size of Image

These are usually based on concave mirror, convex mirror, or convex lens.

Pattern 2: Given Magnification, Find Image or Object Distance

In such questions, students must know whether to use mirror magnification or lens magnification.

Pattern 3: Use Radius of Curvature to Find Focal Length

This is common in mirror-based numericals.

Pattern 4: Identify Mirror or Lens from Image Nature

This tests pure concept clarity.

Pattern 5: Refractive Index and Speed of Light

These are usually short direct formula-based questions.

Common Mistakes Students Make in Light Numericals

Mistake 1: Wrong Sign of Focal Length

Students often forget that:

• concave mirror has negative focal length
• convex mirror has positive focal length
• convex lens has positive focal length
• concave lens has negative focal length

Mistake 2: Confusing Mirror Formula and Lens Formula

The formulas look similar, but the sign structure is different.

Mistake 3: Using Wrong Magnification Formula

For mirrors:

m = -v/u

For lenses:

m = v/u

Mistake 4: Forgetting Nature of Real and Virtual Images

This causes errors in sign convention and final interpretation.

Mistake 5: Not Reading the Word “Real” Carefully

If the question says the image is real, then magnification is negative in mirrors and the image can be obtained on a screen.

Best Study Strategy for Light Numericals

Students can master Light numericals by following a pattern-based revision method.

Step 1: Revise Sign Convention Daily

Even a small sign error can change the answer completely.

Step 2: Learn Standard Image Nature Rules

Some answers can be identified directly through concept.

Step 3: Solve Mirror and Lens Numericals Separately First

This helps avoid formula confusion.

Step 4: Practise Ray Diagrams Along with Numericals

Ray diagrams help students understand why the image is real, virtual, enlarged, or diminished.

Step 5: Solve Repeated CBSE Patterns

Board-style practice improves speed and confidence.

Quick Formula Revision Sheet for Students

Mirrors

• Mirror formula: 1/f = 1/v + 1/u
• Magnification: m = -v/u
• Focal length relation: f = R/2

Lenses

• Lens formula: 1/f = 1/v – 1/u
• Magnification: m = v/u

Refraction

• Refractive index: n = speed of light in vacuum / speed of light in medium

Practice Questions for Revision

Important Questions to Solve

• A concave mirror has a focal length 20 cm. Find the image position for an object at 30 cm.
• A convex lens of focal length 15 cm forms an image on a screen. Find the image position if the object is at 30 cm.
• A concave lens forms an image at 12 cm. Find the object position if focal length is 18 cm.
• Light enters from air into water of refractive index 1.33. Find the speed of light in water.
• A mirror always forms a diminished and erect image. Identify the mirror and explain why.

FAQs

Q1. Why are Light numericals important for Class 10 CBSE Boards?

Light numericals are important because this chapter carries high weightage and frequently includes board questions based on mirrors, lenses, magnification, and refraction.

Q2. What is the most common mistake students make in Light numericals?

The most common mistake is using the wrong sign convention for focal length, object distance, or image distance.

Q3. How do I know whether to use mirror formula or lens formula?

If the question involves a mirror, use 1/f = 1/v + 1/u. If it involves a lens, use 1/f = 1/v – 1/u.

Q4. What is the difference between magnification formula for mirrors and lenses?

For mirrors, magnification is m = -v/u. For lenses, magnification is m = v/u.

Q5. Which mirror always forms a virtual, erect, and diminished image?

A convex mirror always forms a virtual, erect, and diminished image for every object position.

Q6. Which lens always forms a virtual, erect, and diminished image?

A concave lens always forms a virtual, erect, and diminished image.

Q7. What happens when an object is placed at the focus of a convex lens?

When an object is placed at the focus of a convex lens, the image is formed at infinity.

Q8. How can I improve in Light numericals quickly?

You can improve quickly by revising sign convention daily, memorising formulas, solving repeated board-style questions, and connecting numericals with ray diagrams.

Conclusion

Light is one of the most important and scoring chapters in Class 10 Physics, especially for CBSE board exams. The chapter becomes much easier when students master sign convention, revise formulas clearly, and practise standard numerical patterns from mirrors, lenses, and refraction. Most errors in this chapter do not come from lack of intelligence. They come from small sign mistakes, wrong formula choice, or weak interpretation of image nature.

The best way to score well in Light numericals is to combine concept clarity with repeated practice. At Deeksha Vedantu, we always encourage students to revise from basic to advanced numerical patterns so that when board questions appear, they feel familiar rather than stressful.