Light Reflection and Refraction — Class 10 Physics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 10 Physics chapter "Light Reflection and Refraction" — 10 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 10 Physics chapter "Light Reflection and Refraction" — 10 important questions with detailed answers for CB…
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NCERT Solutions for Class 10 Physics — Light Reflection and Refraction. Step-by-step answers to all 10 textbook questions from this chapter, written for CBSE board preparation and free to use.
Light Reflection and Refraction — All 10 Questions Solved
Q1. Differentiate between a real image and a virtual image. State two uses of a concave mirror.
A real image is formed when light rays actually converge at a point after reflection or refraction. It can be obtained on a screen. Real images are always inverted.
A virtual image is formed when light rays appear to diverge from a point after reflection or refraction. It cannot be obtained on a screen. Virtual images are always erect.
Two uses of a concave mirror are:
1. Shaving mirrors: To see a larger image of the face for a close shave.
2. Dentists' mirrors: To see enlarged images of the teeth.
Q2. An object is placed 20 cm in front of a concave mirror of focal length 15 cm. Find the position, nature, and magnification of the image.
Given:
Object distance, u = -20 cm (according to New Cartesian Sign Convention, object is always to the left)
Focal length of concave mirror, f = -15 cm (focal length of concave mirror is negative)
We use the mirror formula:
1/v + 1/u = 1/f
Substitute the given values:
1/v + 1/(-20) = 1/(-15)
1/v - 1/20 = -1/15
1/v = 1/20 - 1/15
To subtract the fractions, find a common denominator, which is 60:
1/v = (3 - 4) / 60
1/v = -1/60
v = -60 cm
The image is formed at 60 cm in front of the mirror (the negative sign indicates it's on the same side as the object).
Now, let's find the magnification (m):
m = -v/u
m = -(-60 cm) / (-20 cm)
m = 60 / (-20)
m = -3
Position of the image: 60 cm in front of the mirror.
Nature of the image: Since v is negative, the image is real. Since m is negative, the image is inverted. Since |m| > 1, the image is magnified (3 times the size of the object).
Q3. Define the absolute refractive index of a medium. State Snell's Law of refraction. What happens to the speed of light when it passes from a rarer medium to a denser medium?
Absolute Refractive Index (n): It is defined as the ratio of the speed of light in vacuum (or air) to the speed of light in the given medium.
Mathematically, n = c/v, where 'c' is the speed of light in vacuum (3 x 10^8 m/s) and 'v' is the speed of light in the medium.
Snell's Law of Refraction: It states that for a given pair of media and for light of a given colour, the ratio of the sine of the angle of incidence (sin i) to the sine of the angle of refraction (sin r) is constant.
Mathematically, (sin i) / (sin r) = constant (which is the refractive index of the second medium with respect to the first).
When light passes from a rarer medium (like air) to a denser medium (like water or glass), its speed decreases. This decrease in speed causes the light ray to bend towards the normal.
Q4. The refractive index of diamond is 2.42. What is the meaning of this statement? If the speed of light in vacuum is 3 x 10^8 m/s, calculate the speed of light in diamond.
The statement 'the refractive index of diamond is 2.42' means that the speed of light in diamond is 2.42 times slower than the speed of light in vacuum (or air). In other words, light travels in vacuum 2.42 times faster than it travels in diamond.
Given:
Refractive index of diamond, n_diamond = 2.42
Speed of light in vacuum, c = 3 x 10^8 m/s
We know the formula for refractive index:
n = c / v
where v is the speed of light in the medium.
Rearranging the formula to find v:
v = c / n
v = (3 x 10^8 m/s) / 2.42
v ≈ 1.24 x 10^8 m/s
Therefore, the speed of light in diamond is approximately 1.24 x 10^8 m/s.
Q5. Draw a neat ray diagram to show the formation of the image of an object placed between F and C of a concave mirror. State the characteristics of the image formed.
Ray Diagram:
(Imagine a diagram here with the following elements)
1. Concave Mirror: Draw a concave mirror with its pole (P), principal axis, principal focus (F), and centre of curvature (C).
2. Object: Place an object (e.g., an arrow AB) between F and C on the principal axis.
3. Ray 1: A ray of light from point A, parallel to the principal axis, passes through the principal focus (F) after reflection.
4. Ray 2: A ray of light from point A, passing through the principal focus (F), becomes parallel to the principal axis after reflection.
(Alternatively, a ray passing through C reflects back along the same path, or a ray incident at P reflects symmetrically).
5. Image Formation: The point where the two reflected rays intersect (or appear to intersect) forms the image point A'. Draw a perpendicular from A' to the principal axis to get the full image A'B'. The image will be formed beyond C.
Characteristics of the image formed:
1. Position: Beyond C (Centre of Curvature).
2. Nature: Real and Inverted.
3. Size: Magnified (Larger than the object).
Q6. An object 5 cm high is placed at a distance of 15 cm from a convex lens of focal length 10 cm. Find the position, size, and nature of the image.
Given:
Object height, h_o = +5 cm
Object distance, u = -15 cm (object is placed to the left)
Focal length of convex lens, f = +10 cm (focal length of convex lens is positive)
We use the lens formula:
1/v - 1/u = 1/f
Substitute the given values:
1/v - 1/(-15) = 1/10
1/v + 1/15 = 1/10
1/v = 1/10 - 1/15
To subtract the fractions, find a common denominator, which is 30:
1/v = (3 - 2) / 30
1/v = 1/30
v = +30 cm
Position of the image: The positive sign for v indicates that the image is formed at 30 cm on the other side of the lens (to the right).
Now, let's find the magnification (m) and image size (h_i):
m = h_i / h_o = v / u
Substitute the values of v, u, and h_o:
m = (+30 cm) / (-15 cm)
m = -2
Now, find h_i:
h_i / h_o = m
h_i / 5 cm = -2
h_i = -2 * 5 cm
h_i = -10 cm
Size of the image: 10 cm. The negative sign for h_i indicates that the image is inverted.
Nature of the image: Since v is positive, the image is formed on the opposite side of the lens, so it is real. Since h_i is negative and m is negative, the image is inverted. Since |m| > 1, the image is magnified (2 times the size of the object).
Q7. Define the power of a lens. What is its SI unit? A lens has a power of +2.0 D. What does this mean about the lens type and its focal length?
Power of a lens (P): It is defined as the reciprocal of its focal length (f) in metres. It indicates the degree of convergence or divergence of light rays that a lens can produce.
Mathematically, P = 1/f (where f is in metres).
SI unit: The SI unit of power of a lens is Dioptre (D). One Dioptre is the power of a lens whose focal length is 1 metre.
A lens has a power of +2.0 D means:
1. Lens Type: The positive sign (+) for power indicates that it is a convex lens (a converging lens).
2. Focal Length: We can calculate its focal length:
P = 1/f
2.0 D = 1/f
f = 1 / 2.0 m
f = 0.5 m or 50 cm.
So, the lens is a convex lens with a focal length of 50 cm.
Q8. Two thin lenses, one convex with focal length 20 cm and another concave with focal length 25 cm, are placed in contact. Calculate the power of this combination.
Given:
Focal length of convex lens, f1 = +20 cm (positive for convex lens)
Focal length of concave lens, f2 = -25 cm (negative for concave lens)
First, convert focal lengths to metres for calculating power:
f1 = 20 cm = 0.20 m
f2 = -25 cm = -0.25 m
Now, calculate the power of each lens:
Power of convex lens, P1 = 1 / f1
P1 = 1 / 0.20 D
P1 = +5 D
Power of concave lens, P2 = 1 / f2
P2 = 1 / (-0.25) D
P2 = -4 D
For a combination of thin lenses in contact, the total power (P_total) is the algebraic sum of their individual powers:
P_total = P1 + P2
P_total = (+5 D) + (-4 D)
P_total = +1 D
Therefore, the power of the combination is +1 D.
Q9. Why do automobiles use convex mirrors as rearview mirrors? What is the focal length of a plane mirror?
Convex mirrors are preferred as rearview mirrors in automobiles for the following reasons:
1. Wider Field of View: Convex mirrors diverge light rays, giving a much wider field of view compared to plane mirrors or concave mirrors. This allows the driver to see a larger area of traffic behind them.
2. Always Form Erect Images: Regardless of the object's distance, a convex mirror always forms an erect (upright) image, which is essential for drivers to perceive the orientation of other vehicles correctly.
3. Diminished Images: They form diminished (smaller) images of objects. This helps to fit a larger area into the small mirror, making more objects visible.
Focal length of a plane mirror:
A plane mirror can be considered as a part of a spherical mirror with an infinitely large radius of curvature. Therefore, the focal length of a plane mirror is infinity (∞).
Q10. Draw a ray diagram to show the formation of the image of an object placed at 2F of a convex lens. State the characteristics of the image formed.
Ray Diagram:
(Imagine a diagram here with the following elements)
1. Convex Lens: Draw a convex lens with its optical centre (O), principal axis, principal focus (F1 and F2 on either side), and 2F (2F1 and 2F2 on either side).
2. Object: Place an object (e.g., an arrow AB) at 2F1 on the principal axis.
3. Ray 1: A ray of light from point A, parallel to the principal axis, passes through the principal focus (F2) on the other side of the lens after refraction.
4. Ray 2: A ray of light from point A, passing through the optical centre (O) of the lens, goes undeviated.
5. Image Formation: The point where the two refracted rays intersect forms the image point A'. This intersection point will be exactly at 2F2 on the other side. Draw a perpendicular from A' to the principal axis to get the full image A'B'.
Characteristics of the image formed:
1. Position: At 2F2 (on the other side of the lens).
2. Nature: Real and Inverted.
3. Size: Same size as the object.
Light Reflection and Refraction — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. Which of the following statements is true for the image formed by a plane mirror?
- A. Real and inverted
- B. Virtual and erect
- C. Real and erect
- D. Virtual and inverted
Answer: B. Virtual and erect — A plane mirror always forms a virtual and erect image. This image is laterally inverted and of the same size as the object.
MCQ 2. If the angle of incidence for a light ray striking a plane mirror is 30 degrees, what is the angle of reflection?
- A. 60 degrees
- B. 15 degrees
- C. 30 degrees
- D. 0 degrees
Answer: C. 30 degrees — According to the first law of reflection, the angle of incidence is always equal to the angle of reflection.
MCQ 3. A spherical mirror whose reflecting surface is curved outwards is called a:
- A. Concave mirror
- B. Convex mirror
- C. Plane mirror
- D. Cylindrical mirror
Answer: B. Convex mirror — A convex mirror has its reflecting surface bulging outwards, causing light rays to diverge after reflection.
MCQ 4. Where should an object be placed in front of a concave mirror to obtain a real, inverted, and same-sized image?
- A. At the focus
- B. Between the pole and focus
- C. At the center of curvature
- D. Beyond the center of curvature
Answer: C. At the center of curvature — When an object is placed at the center of curvature (C) of a concave mirror, the image formed is real, inverted, and of the same size, located at C.
MCQ 5. Which phenomenon is responsible for the bending of light when it passes from one transparent medium to another?
- A. Reflection
- B. Refraction
- C. Dispersion
- D. Scattering
Answer: B. Refraction — Refraction is the phenomenon of bending of light as it travels from one medium to another due to a change in its speed.
MCQ 6. The power of a lens is measured in which SI unit?
- A. Meters
- B. Centimeters
- C. Dioptres
- D. Watts
Answer: C. Dioptres — The SI unit for the power of a lens is dioptre (D), which is equivalent to m⁻¹.
MCQ 7. A convex lens is also commonly known as a:
- A. Diverging lens
- B. Converging lens
- C. Cylindrical lens
- D. Plano-concave lens
Answer: B. Converging lens — A convex lens converges parallel rays of light after refraction, hence it is called a converging lens.
MCQ 8. Which of the following mirrors always forms a virtual, erect, and diminished image?
- A. Plane mirror
- B. Concave mirror
- C. Convex mirror
- D. Cylindrical mirror
Answer: C. Convex mirror — A convex mirror always forms a virtual, erect, and diminished image, regardless of the object's position.
MCQ 9. When light passes from a denser medium to a rarer medium, it bends:
- A. Towards the normal
- B. Away from the normal
- C. Along the normal
- D. Does not bend
Answer: B. Away from the normal — When light travels from a denser medium to a rarer medium, its speed increases, causing it to bend away from the normal.
MCQ 10. When an object is placed at 2F of a convex lens, the image is formed:
- A. At F
- B. At 2F
- C. Between F and 2F
- D. Beyond 2F
Answer: B. At 2F — For a convex lens, when an object is placed at 2F, the image formed is real, inverted, and of the same size, located at 2F on the other side of the lens.
MCQ 11. The focal length of a spherical mirror is 15 cm. What is its radius of curvature?
- A. 7.5 cm
- B. 15 cm
- C. 30 cm
- D. 45 cm
Answer: C. 30 cm — For spherical mirrors, the radius of curvature (R) is twice its focal length (f), i.e., R = 2f. So, R = 2 * 15 cm = 30 cm.
MCQ 12. An object is placed 10 cm in front of a concave mirror of focal length 20 cm. The image formed will be:
- A. Real, inverted, enlarged
- B. Virtual, erect, enlarged
- C. Real, inverted, diminished
- D. Virtual, erect, diminished
Answer: B. Virtual, erect, enlarged — For a concave mirror, when an object is placed between the pole (P) and the principal focus (F), the image formed is virtual, erect, and enlarged, located behind the mirror.
MCQ 13. A full-length image of a distant tall building can definitely be seen by using a:
- A. Concave mirror
- B. Convex mirror
- C. Plane mirror
- D. Both concave and plane mirror
Answer: B. Convex mirror — A convex mirror always forms a virtual, erect, and diminished image, allowing it to provide a wider field of view and show full-length images of distant objects.
MCQ 14. The refractive index of water is 1.33 and that of glass is 1.5. In which medium does light travel faster?
- A. Water
- B. Glass
- C. Both same
- D. Cannot be determined
Answer: A. Water — The speed of light is inversely proportional to the refractive index of the medium (v = c/n). Since water has a lower refractive index than glass, light travels faster in water.
MCQ 15. A ray of light passes from air to glass. It bends:
- A. Towards the normal
- B. Away from the normal
- C. Parallel to the normal
- D. Does not bend
Answer: A. Towards the normal — When light travels from a rarer medium (air) to a denser medium (glass), its speed decreases, causing it to bend towards the normal.
MCQ 16. A student measures the power of a lens as +2.5 D. What is the nature and focal length of the lens?
- A. Concave, 40 cm
- B. Convex, 40 cm
- C. Concave, 25 cm
- D. Convex, 25 cm
Answer: B. Convex, 40 cm — A positive power indicates a convex (converging) lens. Focal length f = 1/P = 1/2.5 D = 0.4 m = 40 cm.
MCQ 17. If the speed of light in vacuum is 'c' and in a medium is 'v', then the absolute refractive index of the medium (n) is given by:
- A. n = c * v
- B. n = c / v
- C. n = v / c
- D. n = 1 / (c * v)
Answer: B. n = c / v — The absolute refractive index of a medium is defined as the ratio of the speed of light in vacuum (c) to the speed of light in that medium (v).
MCQ 18. A student wishes to project the image of a candle flame on a screen using a spherical mirror. Which type of mirror should they use?
- A. Plane mirror
- B. Convex mirror
- C. Concave mirror
- D. Any mirror can be used
Answer: C. Concave mirror — Only a concave mirror can form real images, which can be projected onto a screen. Convex mirrors and plane mirrors form virtual images.
MCQ 19. The phenomenon of lateral displacement is observed when light passes through a:
- A. Spherical mirror
- B. Spherical lens
- C. Rectangular glass slab
- D. Prism
Answer: C. Rectangular glass slab — Lateral displacement is the perpendicular shift in the path of light as it emerges from a rectangular glass slab, parallel to the incident ray.
MCQ 20. An object is placed in front of a concave lens. The image formed is always:
- A. Real, inverted, and diminished
- B. Virtual, erect, and enlarged
- C. Virtual, erect, and diminished
- D. Real, erect, and diminished
Answer: C. Virtual, erect, and diminished — A concave lens always forms a virtual, erect, and diminished image, regardless of the object's position.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.