Human Eye numericals are one of the most important scoring areas in Class 10 Physics. Even though this is a comparatively short chapter, students often lose marks because of sign errors, incorrect identification of the defect of vision, confusion between near point and far point, or wrong unit conversion while calculating power. That is why this topic needs structured revision.
In this chapter, students do not just study the human eye theoretically. They also learn how to solve numerical problems related to focal length, power of lens, myopia, hypermetropia, corrective lenses, and image formation. Once the logic of these questions becomes clear, most board-style problems become much easier.
At Deeksha Vedantu, we always guide students to focus on the pattern behind numericals instead of memorising final answers. When students clearly know what to take as u, what to take as v, and how to apply the sign convention, Human Eye numericals become much more manageable.
Why Human Eye Numericals Are Important in Class 10 Physics
This topic is frequently tested because it checks both conceptual understanding and formula application.
What Students Are Usually Asked
- identify the defect of vision
- name the corrective lens
- calculate focal length
- calculate power of lens
- solve near point and far point based questions
- explain the causes of myopia or hypermetropia
Many of these are repeated board-style question types. Once students practise them properly, they can solve a large number of exam questions with confidence.
Core Formulas Used in Human Eye Numericals
Before solving any numerical, students must know the formulas clearly.
| Formula area | Formula |
| Lens formula | 1/f = 1/v – 1/u |
| Power of lens | P = 1/f |
| Magnification of lens | m = v/u |
| Height relation | m = hᵢ/hₒ |
Important Unit Rule
While calculating power, focal length must always be taken in metres.
Sign Convention You Must Remember
A large number of mistakes in Human Eye numericals happen because students forget the sign convention.
Quick Sign Convention Table
| Quantity | Sign rule |
| Object distance u | Usually negative |
| Focal length of convex lens | Positive |
| Focal length of concave lens | Negative |
| Image distance v in corrective lens questions | Usually negative because the image formed is virtual |
Why v Is Usually Negative in Corrective Lens Questions
In near point and far point problems, the corrective lens forms a virtual image for the defective eye. According to sign convention, this image distance is taken as negative.
This one idea is extremely important for exam success.
How to Identify the Defect of Vision
Before solving many numericals, students must first identify whether the person is suffering from myopia or hypermetropia.
Defect of Vision Quick Revision Table
| Defect of vision | What the person can see clearly | What the person cannot see clearly | Corrective lens | Sign of focal length and power |
| Myopia | Nearby objects | Distant objects | Concave lens | Negative |
| Hypermetropia | Distant objects | Nearby objects | Convex lens | Positive |
Myopia
A person with myopia can see nearby objects clearly but cannot see distant objects clearly.
Corrective Lens for Myopia
Myopia is corrected using a concave lens.
Common Causes of Myopia
- elongation of the eyeball
- decrease in the focal length of the eye lens
Hypermetropia
A person with hypermetropia can see distant objects clearly but cannot see nearby objects clearly.
Corrective Lens for Hypermetropia
Hypermetropia is corrected using a convex lens.
Common Causes of Hypermetropia
- shortening of the eyeball
- increase in the focal length of the eye lens
Quick Rules for Near Point and Far Point Questions
Students often get confused in these problems, but the rule is actually simple.
Near Point and Far Point Table
| Question type | Object distance u | Image distance v |
| Near point correction | Usually the normal near point, such as -25 cm | Defective near point with negative sign |
| Far point correction | Object at infinity, so u = -∞ | Defective far point with negative sign |
Near Point Questions
If the near point of a defective eye is given, that value becomes the image distance for the corrective lens.
For example:
- near point = 50 cm
- so v = -50 cm
If the object is to be seen clearly at 25 cm, then:
- u = -25 cm
Far Point Questions
If the far point of a defective eye is given, that value becomes the image distance for the corrective lens.
For example:
- far point = 1.5 m
- so v = -1.5 m
For a distant object, object distance is taken as infinity.
So:
- u = -∞
In practice:
1/∞ = 0
Step-by-Step Method to Solve Human Eye Numericals
Students should follow the same structure in every question.
Step 1: Write What Is Given
List the values of:
- focal length
- object distance
- image distance
- power
- near point or far point
Step 2: Apply the Correct Sign
Before solving, check whether the value should be positive or negative.
Step 3: Identify What You Need to Find
Decide whether the question asks for:
- defect of vision
- nature of lens
- focal length
- power
- magnification
Step 4: Use the Correct Formula
Apply the lens formula or power formula carefully.
Step 5: Convert Unit if Needed
If focal length is in centimetres, convert it into metres before calculating power.
This is one of the most common board-exam mistakes.
Solved Numerical 1: Lens with Focal Length 10 cm
A lens has focal length 10 cm. Find the nature of the lens and its power.
Given
- f = 10 cm
Step 1: Identify Nature of Lens
Since focal length is positive, the lens is a convex lens.
Step 2: Convert Into Metres
10 cm = 0.1 m
Step 3: Calculate Power
P = 1/f = 1/0.1 = 10 D
Answer
- Nature of lens: Convex lens
- Power of lens: +10 D
Solved Numerical 2: Magnification When Object Is at 20 cm
A convex lens has focal length 10 cm. An object is placed 20 cm from the optical centre. Find the sign of magnification.
Given
- f = +10 cm
- u = -20 cm
Step 1: Use Lens Formula
1/f = 1/v – 1/u
1/10 = 1/v – (1/-20)
1/10 = 1/v + 1/20
1/v = 1/10 – 1/20 = 1/20
So, v = 20 cm
Step 2: Calculate Magnification
m = v/u = 20/(-20) = -1
Answer
Magnification is negative.
This means the image formed is real and inverted.
Solved Numerical 3: Near Point Is 50 cm
The near point of a person is 50 cm. Find the nature and power of the corrective lens required to see an object clearly at 25 cm.
Given
- v = -50 cm
- u = -25 cm
Step 1: Use Lens Formula
1/f = 1/v – 1/u
1/f = 1/(-50) – 1/(-25)
1/f = -1/50 + 1/25
1/f = 1/50
So, f = +50 cm
Step 2: Identify Nature of Lens
Since focal length is positive, the corrective lens is convex.
Step 3: Calculate Power
50 cm = 0.5 m
P = 1/0.5 = +2 D
Answer
- Nature of lens: Convex lens
- Power: +2 D
This is a standard hypermetropia correction question.
Solved Numerical 4: Power Is -0.5 Dioptre
A student needs spectacles of power -0.5 D. Name the defect of vision and find the nature and focal length of the lens.
Given
- P = -0.5 D
Step 1: Identify Lens Nature
Since power is negative, focal length is also negative.
A negative focal length means the lens is concave.
Step 2: Identify Defect of Vision
Concave lens is used to correct myopia.
Step 3: Calculate Focal Length
f = 1/P = 1/(-0.5) = -2 m
Answer
- Defect of vision: Myopia
- Nature of lens: Concave lens
- Focal length: -2 m
Solved Numerical 5: Power Is -2.5 Dioptre
A person wears spectacles of power -2.5 D. Find the defect of vision and focal length of the corrective lens.
Given
- P = -2.5 D
Step 1: Identify Defect
Negative power indicates a concave lens.
A concave lens is used for myopia.
Step 2: Calculate Focal Length
f = 1/P = 1/(-2.5) = -0.4 m
Answer
- Defect of vision: Myopia
- Focal length: -0.4 m
Solved Numerical 6: Person Can See Distant Bus but Not Read a Book
A person can read the number plate of a distant bus clearly, but finds difficulty in reading a nearby book. Identify the defect of vision and the corrective lens required.
Understanding the Situation
If a person can see distant objects clearly but cannot see nearby objects clearly, the person has hypermetropia.
Corrective Lens
Hypermetropia is corrected with a convex lens.
Answer
- Defect of vision: Hypermetropia
- Corrective lens: Convex lens
Solved Numerical 7: Near Point Is 75 cm
The near point of a person suffering from hypermetropia is 75 cm. Calculate the focal length and power of the lens required to read a newspaper kept at 25 cm.
Given
- v = -75 cm
- u = -25 cm
Step 1: Use Lens Formula
1/f = 1/(-75) – 1/(-25)
1/f = -1/75 + 1/25
1/f = 2/75
So,
f = 75/2 = 37.5 cm
Step 2: Convert to Metres
37.5 cm = 0.375 m
Step 3: Calculate Power
P = 1/0.375 ≈ +2.67 D
Answer
- Focal length: +37.5 cm
- Power: approximately +2.67 D
Solved Numerical 8: Near Point of a Farsighted Person Is 40 cm
The near point of a farsighted person is 40 cm. He wants to reduce it to 25 cm using spectacles. Find the power and nature of the lens used.
Given
- v = -40 cm
- u = -25 cm
Step 1: Use Lens Formula
1/f = 1/(-40) – 1/(-25)
1/f = -1/40 + 1/25
Taking LCM:
1/f = 3/200
So,
f = 200/3 cm ≈ 66.67 cm
Step 2: Convert to Metres
66.67 cm ≈ 0.6667 m
Step 3: Calculate Power
P = 1/0.6667 ≈ +1.5 D
Step 4: Identify Lens Nature
Since power is positive, the lens is convex.
Answer
- Power: approximately +1.5 D
- Nature of lens: Convex lens
Solved Numerical 9: Far Point of a Short-Sighted Person Is 1.5 m
The far point of a short-sighted person is 1.5 m. Find the focal length, power, and nature of the remedial lens.
Given
- v = -1.5 m
- u = -∞
Step 1: Use Lens Formula
1/f = 1/v – 1/u
1/f = 1/(-1.5) – 0
So,
f = -1.5 m
Step 2: Calculate Power
P = 1/(-1.5) ≈ -0.67 D
Step 3: Identify Lens Nature
Negative focal length means the remedial lens is concave.
Answer
- Focal length: -1.5 m
- Power: approximately -0.67 D
- Nature of lens: Concave lens
Solved Numerical 10: Concave Lens of Focal Length 10 cm
A person with myopia uses a concave lens of focal length 10 cm. Find its power.
Given
- Concave lens means focal length is negative
- f = -10 cm
Step 1: Convert Into Metres
-10 cm = -0.1 m
Step 2: Calculate Power
P = 1/(-0.1) = -10 D
Answer
Power of the lens is -10 D.
Common Patterns in Human Eye Numericals
If students recognise these question types, solving becomes much faster.
Common Pattern Table
| Pattern type | What to do quickly |
| Focal length given, ask nature and power | Check sign of f, convert into metres, then calculate power |
| Power given, ask defect and lens type | Negative power → concave lens → myopia; positive power → convex lens → hypermetropia |
| Near point given | Take that value as v with negative sign |
| Far point given | Take that value as v with negative sign and use u = -∞ |
| Ask for causes of defect | Link myopia with eyeball elongation or reduced focal length, and hypermetropia with shorter eyeball or increased focal length |
Common Mistakes Students Make in Human Eye Numericals
Forgetting the Sign of Concave Lens
Many students write positive focal length for concave lens. This is incorrect.
Not Converting cm to m
Power can only be calculated correctly when focal length is in metres.
Mixing Up Myopia and Hypermetropia
Remember:
- myopia = difficulty seeing far objects
- hypermetropia = difficulty seeing near objects
Using Wrong Value for v in Corrective Lens Questions
In near point and far point problems, students often choose the wrong image distance. This must be handled carefully.
Board Exam Tips for Human Eye Numericals
Always Write Given, To Find, Formula, Solution
This gives your answer a clean presentation and reduces careless mistakes.
Mark the Sign Before Starting the Calculation
Do not wait until the end. Write sign convention first.
Solve Step by Step
Even if the answer is simple, show each step clearly.
Approximate Only at the End
When decimals come in power values, keep the exact value through the calculation and round off only at the final step.
Important Questions to Practise for Exams
Numerical Practice Questions
- A lens has focal length 20 cm. Find its power and identify its type.
- A student uses spectacles of power -1 D. Name the defect and lens used.
- The near point of a hypermetropic person is 60 cm. Find the power needed to read at 25 cm.
- The far point of a myopic person is 2 m. Find the focal length and power of the corrective lens.
- A person with myopia uses a lens of focal length -50 cm. Find the power.
Conceptual Practice Questions
- Why is a concave lens used to correct myopia?
- Why is a convex lens used to correct hypermetropia?
- Why is sign convention important in lens numericals?
- What happens if focal length is not converted into metres before calculating power?
FAQs
Q1. What is the most important formula for Human Eye numericals in Class 10?
The most important formula is the lens formula, 1/f = 1/v – 1/u, along with the power formula, P = 1/f where focal length is in metres.
Q2. How do I identify whether the defect is myopia or hypermetropia?
If a person cannot see distant objects clearly, the defect is myopia. If a person cannot see nearby objects clearly, the defect is hypermetropia.
Q3. Which lens is used for myopia and hypermetropia?
Myopia is corrected using a concave lens, while hypermetropia is corrected using a convex lens.
Q4. Why is the image distance negative in corrective lens questions?
In such questions, the corrective lens forms a virtual image for the defective eye. According to sign convention, this image distance is taken as negative.
Q5. Why do students lose marks in Human Eye numericals?
Most mistakes happen because of wrong sign convention, incorrect defect identification, failure to convert centimetres to metres, or confusion between near point and far point.
Q6. What is the normal near point and far point of a healthy human eye?
The normal near point is 25 cm and the normal far point is infinity.
Q7. How do I calculate power from focal length?
Use the formula P = 1/f, but make sure the focal length is in metres before calculating the answer in dioptres.
Q8. What are the two main causes of myopia?
The two main causes of myopia are elongation of the eyeball and decrease in the focal length of the eye lens.
Conclusion
Human Eye numericals in Class 10 Physics are highly manageable once students understand the logic of sign convention, defect identification, and formula application. Most board questions come from repeated patterns such as near point correction, far point correction, power calculation, and identifying whether a lens is convex or concave.
The best way to score well in this topic is to practise standard numerical types again and again until the process becomes automatic. At Deeksha Vedantu, we always encourage students to solve these questions with proper steps, because clarity in method leads directly to accuracy in the final answer.
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