Class 10 Light NCERT Solutions (Reflection & Refraction Explained)

Relevant Theory

A lens is an optical device made of a transparent material that refracts (bends) light rays passing through it to form images. The key requirement for any material to be used as a lens is:

It must allow light to pass through it (transparency).
It should have a uniform refractive index to bend light properly.
It must be optically homogeneous for clear image formation.

Solution Roadmap

Step 1: Recall the definition of a lens.
Step 2: Identify the key property required → transparency.
Step 3: Check each option for transparency.
Step 4: Select the material that does NOT allow light to pass.

Step-by-Step Solution

Step 1: A lens must be made of a material that allows light to pass through it (transparent material).

Step 2: Check each option:

Water → Transparent, allows refraction of light ✔
Glass → Transparent, commonly used for lenses ✔
Plastic → Transparent (in optical grade), used in lenses ✔
Clay → Opaque, does not allow light to pass ✖

Step 3: Since clay does not transmit light, it cannot refract light and hence cannot form images.

Final Answer: (d) Clay

Conceptual Ray Interaction

Significance for Exams

Frequently asked conceptual MCQ in board exams (1 mark).
Important for understanding the basic requirement of optical devices.
Helps in competitive exams (NTSE, Olympiads, JEE Foundation) where material properties and optics are tested.
Builds foundation for deeper topics like refraction through lenses and image formation.

Q2

NUMERIC3 marks

The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should be the position of the object?
(a) Between the principal focus and the centre of curvature
(b) At the centre of curvature
(c) Beyond the centre of curvature
(d) Between the pole of the mirror and its principal focus.

Relevant Theory

A concave mirror forms different types of images depending on the position of the object relative to the pole (P), focus (F), and centre of curvature (C).

When the object is placed between pole (P) and focus (F), reflected rays diverge and appear to come from behind the mirror.
Hence, the image formed is virtual, erect, and magnified.
Such images cannot be obtained on a screen because they are formed by apparent intersection of rays.

Solution Roadmap

Step 1: Identify image characteristics → virtual, erect, enlarged.
Step 2: Recall standard image formation cases of concave mirror.
Step 3: Match the given characteristics with correct object position.
Step 4: Select the correct option.

Step-by-Step Solution

Step 1: Given image characteristics:

Nature → Virtual
Orientation → Erect
Size → Enlarged (magnified)

Step 2: Recall concave mirror cases:

Object Position	Nature	Orientation	Size
Between P and F	Virtual	Erect	Enlarged
Between F and C	Real	Inverted	Enlarged
At C	Real	Inverted	Same size
Beyond C	Real	Inverted	Smaller

Step 3: Only the case between pole (P) and focus (F) satisfies all given conditions.

Final Answer: (d) Between the pole of the mirror and its principal focus.

Ray Diagram (Object between P and F)

Significance for Exams

Very important conceptual MCQ for CBSE board exams (frequently repeated).
Forms the basis for ray diagram questions (2–3 marks).
Essential for competitive exams like NTSE, Olympiads, and JEE Foundation.
Helps in understanding real-life applications such as shaving mirrors and makeup mirrors.

Relevant Theory

A convex lens (converging lens) forms different types of images depending on the object position. The size and nature of the image are determined using ray diagrams and magnification.

When an object is placed at twice the focal length (2F₁), the image is formed at 2F₂.
The image formed is real, inverted, and of the same size as the object.
Magnification \( m = \frac{h_i}{h_o} = 1 \) (in magnitude).

Solution Roadmap

Step 1: Identify required image → real and same size.
Step 2: Recall standard convex lens image formation cases.
Step 3: Find the object position that gives magnification = 1.
Step 4: Match with given options.

Step-by-Step Solution

Step 1: Required image properties:

Nature → Real
Size → Same as object

Step 2: Use magnification relation:

\[ m = \frac{h_i}{h_o} \]

Step 3: For same size image:

\[ m = 1 \]

Step 4: From standard convex lens results:

At object position = 2F₁, image forms at 2F₂
Image is real, inverted, and same size

Step 5: Match with options:

Correct option is (b) At twice the focal length.

Final Answer: (b) At twice the focal length

Ray Diagram (Object at 2F₁)

Significance for Exams

Very common CBSE board MCQ and short-answer concept.
Important for ray diagram drawing questions (3 marks).
Helps in solving numerical problems using lens formula and magnification.
Frequently tested in NTSE, Olympiads, and JEE Foundation exams.

Relevant Theory

In optics, the New Cartesian Sign Convention is used to determine the nature of mirrors and lenses. According to this convention:

All distances are measured from the pole (mirror) or optical centre (lens).
Distances measured in the direction of incident light are taken as positive.
Distances measured opposite to the direction of incident light are taken as negative.

For Mirrors:

Concave mirror → focal length is negative
Convex mirror → focal length is positive

For Lenses:

Convex lens (converging) → focal length is positive
Concave lens (diverging) → focal length is negative

Solution Roadmap

Step 1: Note the given focal length (–15 cm).
Step 2: Apply sign convention separately for mirror and lens.
Step 3: Identify the nature based on sign.
Step 4: Match with the correct option.

Step-by-Step Solution

Step 1: Given focal length:

\[ f = -15 \text{ cm} \]

Step 2: For spherical mirror:

Negative focal length ⇒ mirror is concave

Step 3: For thin spherical lens:

Negative focal length ⇒ lens is concave (diverging lens)

Step 4: Combine both results:

Mirror → Concave
Lens → Concave

Final Answer: (a) both concave.

Conceptual Representation

Significance for Exams

Very important MCQ based on sign convention in CBSE board exams.
Forms the foundation for solving numerical problems using mirror and lens formulas.
Frequently asked in NTSE, Olympiads, and JEE Foundation exams.
Helps avoid common mistakes related to sign convention in optics.

Relevant Theory

The nature of the image formed by a mirror depends on its type:

Plane mirror → Always forms virtual, erect, and same-sized images regardless of object distance.
Convex mirror → Always forms virtual, erect, and diminished images for all object positions.
Concave mirror → Forms:
- Erect image only when object is between pole (P) and focus (F)
- Inverted image for all other positions

Solution Roadmap

Step 1: Identify key condition → image is always erect.
Step 2: Check which mirrors always produce erect images.
Step 3: Eliminate mirrors that sometimes produce inverted images.
Step 4: Choose correct option.

Step-by-Step Solution

Step 1: Given condition:

Image is always erect, irrespective of object distance.

Step 2: Analyze each type of mirror:

Plane mirror: Always forms erect images ✔
Convex mirror: Always forms erect images ✔
Concave mirror: Forms erect image only when object is between P and F, otherwise inverted ✖

Step 3: Eliminate concave mirror because it does not always produce erect images.

Step 4: Remaining possibilities:

Plane mirror
Convex mirror

Final Answer: (d) either plane or convex.

Conceptual Comparison

Significance for Exams

Common conceptual MCQ in CBSE board exams.
Tests understanding of image formation rules for mirrors.
Important for competitive exams like NTSE and Olympiads.
Helps in real-life applications such as rear-view mirrors (convex mirrors).

Relevant Theory

To read small letters clearly, we need a magnified (enlarged) image. This is achieved using a convex lens, also known as a converging lens.

A convex lens produces a virtual, erect, and magnified image when the object is placed between the optical centre (O) and principal focus (F).
A concave lens always produces a virtual but diminished (smaller) image, so it is not useful for magnification.
Magnification increases when the focal length is smaller.

Solution Roadmap

Step 1: Identify requirement → magnified image for small letters.
Step 2: Choose lens type that magnifies → convex lens.
Step 3: Compare focal lengths → smaller focal length gives higher magnification.
Step 4: Select the best option.

Step-by-Step Solution

Step 1: Requirement:

We need to read small letters → image must be magnified.

Step 2: Choose correct type of lens:

Convex lens → produces magnified image ✔
Concave lens → always produces diminished image ✖

Step 3: Compare focal lengths of convex lenses:

Focal length = 50 cm → low magnification
Focal length = 5 cm → high magnification ✔

Step 4: Therefore, the best lens is:

Final Answer: (c) A convex lens of focal length 5 cm.

Ray Diagram (Magnifying Glass)

Significance for Exams

Frequently asked application-based MCQ in CBSE exams.
Important for understanding magnification and optical instruments.
Forms the basis for questions on microscopes and magnifying glasses.
Useful for NTSE, Olympiads, and foundational competitive exams.

Relevant Theory

A concave mirror can form both real and virtual images depending on object position.

When the object is placed between pole (P) and focus (F), the image formed is:
- Virtual
- Erect
- Magnified (larger than object)
The image is formed behind the mirror due to apparent divergence of reflected rays.

Solution Roadmap

Step 1: Identify condition → erect image.
Step 2: Recall position for erect image in concave mirror.
Step 3: Express object distance range using focal length.
Step 4: State nature and size of image.
Step 5: Draw ray diagram.

Step-by-Step Solution

Step 1: Given focal length:

\[ f = -15 \text{ cm} \] (negative sign indicates concave mirror)

Step 2: For erect image in a concave mirror:

Object must be placed between pole (P) and focus (F).

Step 3: Express object distance range:

\[ 0 < |u| < 15 \text{ cm} \]

Using sign convention:

\[ -15 \text{ cm} < u < 0 \]

Step 4: Nature of image:

Virtual
Erect
Formed behind the mirror

Step 5: Size of image:

Image is larger (magnified) than the object.

Final Answer: Object should be placed between P and F (0 < |u| < 15 cm). Image formed is virtual, erect, and larger than the object.

Ray Diagram (Object between P and F)

Significance for Exams

Very important long-answer question for CBSE board exams (3–5 marks with diagram).
Tests conceptual clarity of image formation in concave mirrors.
Ray diagram drawing is frequently evaluated strictly in board marking schemes.
Important for competitive exams like NTSE, Olympiads, and JEE Foundation.

Relevant Theory

The choice of mirror in practical applications depends on how it reflects light:

Concave mirror: Converges parallel rays to a focus and can also produce parallel rays when the source is at focus.
Convex mirror: Diverges light rays and provides a wide field of view with diminished images.

Solution Roadmap

Step 1: Identify requirement of each application.
Step 2: Match with mirror properties (converging/diverging).
Step 3: Justify with ray behavior.

Step-by-Step Solution

(a) Headlights of a Car

Step 1: Requirement → Strong beam of light travelling long distance.

Step 2: Property needed → Parallel rays.

Step 3: A concave mirror produces parallel rays when the light source is placed at its focus.

Answer: Concave Mirror

(b) Side/Rear-View Mirror

Step 1: Requirement → Wide field of view to see more area behind.

Step 2: Property needed → Diverging rays and upright image.

Step 3: A convex mirror provides a wider field of view and forms diminished, erect images.

Answer: Convex Mirror

(c) Solar Furnace

Step 1: Requirement → Concentrate sunlight at one point to generate heat.

Step 2: Property needed → Converging rays.

Step 3: A concave mirror converges parallel rays (sunlight) at its focus.

Answer: Concave Mirror

Conceptual Ray Diagrams

Summary Table

Application	Type of Mirror	Reason
Headlights of a car	Concave	Produces parallel beam for long-distance illumination
Side/rear-view mirror	Convex	Provides wide field of view with erect image
Solar furnace	Concave	Converges sunlight to a focal point for heating

Significance for Exams

Frequently asked application-based question in CBSE board exams (2–3 marks).
Tests conceptual understanding of mirror properties.
Important for NTSE, Olympiads, and competitive exams.
Helps relate physics concepts to real-life applications.

Relevant Theory

A convex lens forms an image by refraction of light rays passing through all parts of the lens. Each portion of the lens contributes to the formation of the entire image.

Even if part of the lens is blocked, the remaining portion still receives light from the whole object.
However, fewer rays pass through → image brightness decreases.
Image size and position remain unchanged.

Solution Roadmap

Step 1: Understand how a lens forms an image.
Step 2: Analyze effect of blocking part of the lens.
Step 3: Predict changes in image formation.
Step 4: Verify through experiment.

Step-by-Step Solution

Step 1: Consider a convex lens forming an image of an object.

Step 2: Cover half of the lens with black paper.

Step 3: Observation:

The image is still complete.
The image appears less bright (fainter).

Step 4: Explanation:

Each part of the lens forms the entire image.
When half the lens is covered, fewer light rays contribute → brightness decreases.
No part of the image disappears because the uncovered portion still receives light from the whole object.

Step 5: Conclusion:

A complete image is formed, but it is less bright.

Experimental Verification

Take a convex lens and form a clear image of an object on a screen.
Now cover half of the lens with black paper.
Observe the image on the screen.
You will notice that the image remains complete but becomes dim.

Significance for Exams

Very important conceptual and experimental question for CBSE board exams (3 marks).
Tests understanding of image formation by lenses.
Frequently asked in practical-based questions.
Important for NTSE and Olympiad exams.

Relevant Theory

Image formation by a thin lens is governed by the lens formula and magnification:

\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]

\[ m = \frac{h'}{h} = \frac{v}{u} \]

Sign convention must be followed strictly.
Negative magnification → inverted image.
|m| < 1 → image is smaller than object.

Solution Roadmap

Step 1: Write given quantities with proper sign.
Step 2: Apply lens formula to find image distance (v).
Step 3: Use magnification to find image height.
Step 4: Determine nature using sign of v and m.

Step-by-Step Solution

Step 1: Given Data

\[ h = 5 \text{ cm}, \quad u = -25 \text{ cm}, \quad f = +10 \text{ cm} \]

Step 2: Apply Lens Formula

\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]

\[ \frac{1}{10} = \frac{1}{v} - \left(\frac{1}{-25}\right) \]

\[ \frac{1}{10} = \frac{1}{v} + \frac{1}{25} \]

\[ \frac{1}{v} = \frac{1}{10} - \frac{1}{25} \]

\[ \frac{1}{v} = \frac{5 - 2}{50} = \frac{3}{50} \]

\[ v = \frac{50}{3} \approx 16.67 \text{ cm} \]

Step 3: Magnification

\[ m = \frac{v}{u} = \frac{50/3}{-25} \]

\[ m = -\frac{2}{3} \]

Step 4: Image Height

\[ m = \frac{h'}{h} \]

\[ -\frac{2}{3} = \frac{h'}{5} \]

\[ h' = -\frac{10}{3} \approx -3.33 \text{ cm} \]

Step 5: Interpretation

v is positive → image is formed on the other side → real image
m is negative → image is inverted
|m| < 1 → image is smaller (diminished)

Final Answer:
Image distance = 16.67 cm
Image height = 3.33 cm (inverted)
Nature = Real, inverted, and diminished

Ray Diagram

Significance for Exams

Very important numerical problem for CBSE board exams (3–4 marks).
Combines lens formula, magnification, and sign convention.
Ray diagram is often required for full marks.
Frequently asked in NTSE, Olympiads, and JEE Foundation exams.

Relevant Theory

For a thin lens, image formation is governed by:

\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]

\[ m = \frac{v}{u} \]

For a concave lens, focal length is negative.
Image formed is always virtual, erect, and diminished.
Image distance (v) is negative (same side as object).

Solution Roadmap

Step 1: Write given values using sign convention.
Step 2: Apply lens formula.
Step 3: Solve for object distance (u).
Step 4: Verify nature using magnification.

Step-by-Step Solution

Step 1: Given Data

\[ f = -15 \text{ cm}, \quad v = -10 \text{ cm} \]

Step 2: Lens Formula

\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]

\[ \frac{1}{-15} = \frac{1}{-10} - \frac{1}{u} \]

Step 3: Rearranging:

\[ \frac{1}{u} = \frac{1}{-10} - \frac{1}{-15} \]

\[ \frac{1}{u} = -\frac{1}{10} + \frac{1}{15} \]

\[ \frac{1}{u} = \frac{-3 + 2}{30} = -\frac{1}{30} \]

\[ u = -30 \text{ cm} \]

Step 4: Magnification

\[ m = \frac{v}{u} = \frac{-10}{-30} = \frac{1}{3} \]

Step 5: Interpretation:

u is negative → object is in front of lens
m is positive → image is erect
|m| < 1 → image is diminished

Final Answer:
Object distance = 30 cm in front of the lens
Image is virtual, erect, and diminished

Ray Diagram (Concave Lens)

Significance for Exams

Important numerical problem for CBSE board exams (3–4 marks).
Tests proper use of sign convention in lens formula.
Ray diagram is essential for full marks.
Frequently asked in NTSE, Olympiads, and foundation-level competitive exams.

Relevant Theory

For spherical mirrors, image formation is governed by the mirror formula:

\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]

Magnification:

\[ m = -\frac{v}{u} \]

Convex mirror → focal length is positive.
Image formed is always virtual, erect, and diminished.
Image distance (v) is positive (behind the mirror).

Solution Roadmap

Step 1: Write given values using sign convention.
Step 2: Apply mirror formula.
Step 3: Solve for image distance (v).
Step 4: Use magnification to determine nature.

Step-by-Step Solution

Step 1: Given Data

\[ u = -10 \text{ cm}, \quad f = +15 \text{ cm} \]

Step 2: Apply Mirror Formula

\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]

\[ \frac{1}{15} = \frac{1}{v} + \left(\frac{1}{-10}\right) \]

\[ \frac{1}{15} = \frac{1}{v} - \frac{1}{10} \]

\[ \frac{1}{v} = \frac{1}{15} + \frac{1}{10} \]

\[ \frac{1}{v} = \frac{2 + 3}{30} = \frac{5}{30} = \frac{1}{6} \]

\[ v = 6 \text{ cm} \]

Step 3: Magnification

\[ m = -\frac{v}{u} = -\frac{6}{-10} = 0.6 \]

Step 4: Interpretation

v is positive → image formed behind the mirror → virtual image
m is positive → image is erect
|m| < 1 → image is diminished (smaller)

Final Answer:
Image position = 6 cm behind the mirror
Nature = Virtual, erect, and diminished

Ray Diagram (Convex Mirror)

Significance for Exams

Frequently asked numerical problem in CBSE board exams (3 marks).
Tests correct use of mirror formula and sign convention.
Ray diagram is often required for full marks.
Important for NTSE, Olympiads, and foundational competitive exams.

Relevant Theory

Magnification for mirrors is defined as:

\[ m = \frac{h'}{h} = -\frac{v}{u} \]

\(m > 0\) → Image is erect (upright)
\(m < 0\) → Image is inverted
\(|m| = 1\) → Image size is equal to object size

Solution Roadmap

Step 1: Interpret sign of magnification.
Step 2: Interpret magnitude of magnification.
Step 3: Conclude image characteristics.

Step-by-Step Explanation

Step 1: Given:

\[ m = +1 \]

Step 2: Interpret the sign:

Positive sign (+) indicates that the image is erect (upright).

Step 3: Interpret the magnitude:

Magnitude = 1 → image height = object height.

Step 4: Additional property of plane mirror:

Image is formed at the same distance behind the mirror as the object is in front.
Image is always virtual.

Final Answer:
Magnification +1 means the image formed is virtual, erect, and of the same size as the object, and it is formed at equal distance behind the mirror.

Conceptual Diagram

Significance for Exams

Frequently asked 1-mark conceptual question in CBSE board exams.
Helps understand sign and magnitude of magnification clearly.
Important for numerical problems involving mirrors.
Useful for NTSE and Olympiad-level conceptual questions.

Relevant Theory

For spherical mirrors:

\[ f = \frac{R}{2} \]

Mirror formula:

\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]

Magnification:

\[ m = -\frac{v}{u} = \frac{h'}{h} \]

Convex mirror → \(f > 0\)
Image is always virtual, erect, and diminished

Solution Roadmap

Step 1: Convert radius to focal length.
Step 2: Apply mirror formula to find image distance.
Step 3: Calculate magnification.
Step 4: Find image height.
Step 5: State nature of image.

Step-by-Step Solution

Step 1: Given Data

\[ \begin{aligned} h &= 5 \text{ cm},\\ u &= -20 \text{ cm},\\ R &= 30 \text{ cm} \end{aligned} \]

Step 2: Find Focal Length

\[ f = \frac{R}{2} = \frac{30}{2} = 15 \text{ cm} \]

Step 3: Apply Mirror Formula

\[ \frac{1}{15} = \frac{1}{v} + \frac{1}{-20} \]

\[ \frac{1}{15} = \frac{1}{v} - \frac{1}{20} \]

\[ \frac{1}{v} = \frac{1}{15} + \frac{1}{20} \]

\[ \frac{1}{v} = \frac{4 + 3}{60} = \frac{7}{60} \]

\[ v = \frac{60}{7} \approx 8.57 \text{ cm} \]

Step 4: Magnification

\[ m = -\frac{v}{u} = -\frac{60/7}{-20} \]

\[ m = \frac{3}{7} \approx 0.43 \]

Step 5: Image Height

\[ \begin{aligned} h' &= m \cdot h \\&= \frac{3}{7} \times 5 \\&= \frac{15}{7} \\&\approx 2.14 \text{ cm} \end{aligned} \]

Step 6: Interpretation

v is positive → image formed behind mirror → virtual
m is positive → image is erect
|m| < 1 → image is diminished

Final Answer:
Image position = 8.57 cm behind the mirror
Image height = 2.14 cm
Nature = Virtual, erect, and diminished

Ray Diagram (Convex Mirror)

Significance for Exams

Very important numerical involving radius and focal length conversion.
Tests mirror formula and magnification together.
Ray diagram is often required for full marks.
Frequently appears in CBSE boards and NTSE exams.

Relevant Theory

For spherical mirrors:

\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]

Magnification:

\[ m = -\frac{v}{u} = \frac{h'}{h} \]

Concave mirror → \(f < 0\)
Real image → formed on screen
Negative magnification → inverted image

Solution Roadmap

Step 1: Write given values with sign convention.
Step 2: Apply mirror formula to find image distance.
Step 3: Determine screen position.
Step 4: Calculate magnification and image height.
Step 5: State nature of image.

Step-by-Step Solution

Step 1: Given Data

\[ h = 7 \text{ cm}, \quad u = -27 \text{ cm}, \quad f = -18 \text{ cm} \]

Step 2: Apply Mirror Formula

\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]

\[ \frac{1}{-18} = \frac{1}{v} + \frac{1}{-27} \]

\[ \frac{1}{-18} = \frac{1}{v} - \frac{1}{27} \]

\[ \frac{1}{v} = \frac{1}{-18} + \frac{1}{27} \]

\[ \frac{1}{v} = -\frac{1}{18} + \frac{1}{27} \]

\[ \frac{1}{v} = \frac{-3 + 2}{54} = -\frac{1}{54} \]

\[ v = -54 \text{ cm} \]

Step 3: Screen Position

v is negative → image is formed in front of mirror → real image
Screen should be placed at 54 cm in front of the mirror

Step 4: Magnification

\[ m = -\frac{v}{u} = -\frac{-54}{-27} \]

\[ m = -2 \]

Step 5: Image Height

\[ m = \frac{h'}{h} \]

\[ -2 = \frac{h'}{7} \]

\[ h' = -14 \text{ cm} \]

Step 6: Interpretation

v negative → real image
m negative → inverted image
|m| > 1 → magnified image

Final Answer:
Screen position = 54 cm in front of mirror
Image size = 14 cm
Nature = Real, inverted, and magnified

Ray Diagram (Concave Mirror)

Significance for Exams

Very important long numerical (4–5 marks) in CBSE board exams.
Combines mirror formula, magnification, and practical concept of screen placement.
Ray diagram is essential for full scoring.
Frequently asked in NTSE and Olympiad exams.

Relevant Theory

The power of a lens is defined as:

\[ P = \frac{1}{f} \]

\(P\) is in dioptre (D) and \(f\) is in metre (m).
\(P > 0\) → Convex lens (converging)
\(P < 0\) → Concave lens (diverging)

Solution Roadmap

Step 1: Write given power.
Step 2: Use relation \(P = 1/f\).
Step 3: Convert focal length into cm.
Step 4: Determine type from sign.

Step-by-Step Solution

Step 1: Given Data

\[ P = +2.0 \text{ D} \]

Step 2: Use Formula

\[ P = \frac{1}{f} \]

\[ 2 = \frac{1}{f} \]

\[ f = \frac{1}{2} = 0.5 \text{ m} \]

Step 3: Convert into cm

\[ f = 0.5 \times 100 = 50 \text{ cm} \]

Step 4: Determine Type

Focal length is positive → lens is convex.

Final Answer:
Focal length = 50 cm
Type of lens = Convex lens

Conceptual Representation

Significance for Exams

Very common 1–2 mark numerical in CBSE board exams.
Tests understanding of power–focal length relation.
Sign of power is frequently used to identify lens type.
Important for competitive exams like NTSE and Olympiads.

Relevant Theory

The relation between power and focal length is:

\[ P = \frac{1}{f} \]

Power \(P\) is in dioptre (D), focal length \(f\) in metre (m).
\(P > 0\) → Converging (convex) lens.
\(P < 0\) → Diverging (concave) lens.

Solution Roadmap

Step 1: Note given power.
Step 2: Apply \(P = 1/f\).
Step 3: Convert focal length into cm.
Step 4: Determine type using sign.

Step-by-Step Solution

Step 1: Given Data

\[ P = +1.5 \text{ D} \]

Step 2: Use Formula

\[ P = \frac{1}{f} \]

\[ 1.5 = \frac{1}{f} \]

\[ f = \frac{1}{1.5} \]

\[ f = \frac{2}{3} \text{ m} \]

\[ f \approx 0.67 \text{ m} \]

Step 3: Convert into cm

\[ f = 0.67 \times 100 = 66.7 \text{ cm} \approx 67 \text{ cm} \]

Step 4: Determine Type

Positive focal length → lens is converging (convex).

Final Answer:
Focal length ≈ 0.67 m (≈ 67 cm)
Type of lens = Converging (Convex lens)

Conceptual Representation

Significance for Exams

Frequently asked medical application question in CBSE exams.
Important for understanding corrective lenses (vision defects).
Tests relation between power and focal length.
Relevant for NTSE, Olympiads, and NEET foundation.

Light - Reflection and Refraction

Relevant Theory

Solution Roadmap

Step-by-Step Solution

Conceptual Ray Interaction

Significance for Exams

Relevant Theory

Solution Roadmap

Step-by-Step Solution

Ray Diagram (Object between P and F)

Significance for Exams

Relevant Theory

Solution Roadmap

Step-by-Step Solution

Ray Diagram (Object at 2F₁)

Significance for Exams

Relevant Theory

Solution Roadmap

Step-by-Step Solution

Conceptual Representation

Significance for Exams

Relevant Theory

Solution Roadmap

Step-by-Step Solution

Conceptual Comparison

Significance for Exams

Relevant Theory

Solution Roadmap

Step-by-Step Solution

Ray Diagram (Magnifying Glass)

Significance for Exams

Relevant Theory

Solution Roadmap

Step-by-Step Solution

Ray Diagram (Object between P and F)

Significance for Exams

Relevant Theory

Solution Roadmap

Step-by-Step Solution

Conceptual Ray Diagrams

Summary Table

Significance for Exams

Relevant Theory

Solution Roadmap

Step-by-Step Solution

Experimental Verification

Significance for Exams

Relevant Theory

Solution Roadmap

Step-by-Step Solution

Ray Diagram

Significance for Exams

Relevant Theory

Solution Roadmap

Step-by-Step Solution

Ray Diagram (Concave Lens)

Significance for Exams

Relevant Theory

Solution Roadmap

Step-by-Step Solution

Ray Diagram (Convex Mirror)

Significance for Exams

Relevant Theory

Solution Roadmap

Step-by-Step Explanation

Conceptual Diagram

Significance for Exams

Relevant Theory

Solution Roadmap

Step-by-Step Solution

Ray Diagram (Convex Mirror)

Significance for Exams

Relevant Theory

Solution Roadmap

Step-by-Step Solution

Ray Diagram (Concave Mirror)

Significance for Exams

Relevant Theory

Solution Roadmap

Step-by-Step Solution