Looking for accurate SEBA Class 10 Science Chapter 10 (Light – Reflection and Refraction) textual solutions for the 2026–27 academic year? Access step-by-step textbook exercise answers, ray diagram explanations, mirror formula and lens formula calculations, magnification rules, and sign convention notes to help you score top marks in your HSLC board exams.
Blue question
1. Define the principal focus of a concave mirror.
Answer: The principal focus of a concave mirror is a point on its principal axis where rays of light coming parallel to the principal axis meet after reflection from the mirror.
2. The radius of curvature of a spherical mirror is 20 cm. What is its focal length?
Answer: For a spherical mirror, f = R / 2
Given, Radius of curvature (R) = 20 cm f = 20 / 2 = 10 cm
Therefore, the focal length of the mirror is 10 cm.
3. Name a mirror that can give an erect and enlarged image of an object.
Answer: A concave mirror can produce an erect and enlarged image when the object is placed between the pole and the principal focus of the mirror.
4. Why do we prefer a convex mirror as a rear-view mirror in vehicles?
Answer: A convex mirror is used as a rear-view mirror because it shows a larger area behind the vehicle. It forms an erect image and helps the driver see more of the road, making driving safer.
5. Find the focal length of a convex mirror whose radius of curvature is 32 cm.
Answer:
For a spherical mirror, the focal length f is given by:
f = R / 2
If R = 32 cm, then:
f = 32 / 2 = 16 cm
(For a convex mirror, this focus is virtual and located behind the mirror; in sign-convention problems, it is often taken as f = -16 cm.)
SEBA Class 10 Science Chapter 10 Solutions
6. A concave mirror produces three times magnified (enlarged) real image of an object placed at 10 cm in front of it. Where is the image located?
Answer:
Linear magnification (m) is given by:
m = (image distance) / (object distance)
m = v / u
Given |m| = 3 and object distance u = 10 cm,
then image distance v = 3 × 10 = 30 cm.
Thus, the real image is located 30 cm in front of the mirror (on the same side as the object).
(Optional: To find the mirror’s focal length)
1/f = 1/u + 1/v
1/f = 1/10 + 1/30 = 4/30
f = 7.5 cm
7. A ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?
Answer: When a ray of light travels from air into water, it bends towards the normal. This happens because water is a denser medium than air, so the speed of light decreases when it enters water.
8. Light enters from air to glass having refractive index 1.50. What is the speed of light in the glass? (Speed of light in vacuum = 3 x 10^8 m/s).
Answer:
Speed of light in glass is given by:
v = c / n
Given:
c = 3 × 10⁸ m/s (speed of light in vacuum)
n = 1.50 (refractive index of glass)
Then:
v = (3 × 10⁸) / 1.50 = 2.0 × 10⁸ m/s
9. From a typical refractive-index table, which medium has the highest optical density and which has the lowest?
Answer: The optical density depends on the refractive index of a medium.
Diamond has the highest optical density because it has the highest refractive index (about 2.42).
Air has the lowest optical density because its refractive index is nearly 1.00.
10. You are given kerosene, turpentine and water. In which of these does the light travel fastest?
Answer: Light travels fastest in the medium with the lowest refractive index.
Among water (1.33), kerosene (1.44), and turpentine (1.47), water has the lowest refractive index.
Therefore, light travels fastest in water.
Class 10 Science Chapter 10 Question Answer SEBA 2026-27
11. The refractive index of diamond is 2.42. What is the meaning of this statement?
Answer: It means light travels 2.42 times slower in diamond than in vacuum.
The speed of light in diamond is approximately (1.24 \times 10^8) m/s.
This shows that diamond has a very high refractive index and bends light strongly.
12. Define 1 dioptre of power of a lens.
Answer:
1 dioptre (1 D) is the power of a lens whose focal length is 1 metre.
Power P (in dioptres) = 1 / f (in metres).
13. A convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle placed in front of the convex lens if the image is equal to the size of the object? Also, find the power of the lens.
Answer:
Diamond has the largest refractive index, n ≈ 2.42
Air (or vacuum) has refractive index ≈ 1.00
(Higher refractive index ⇒ higher optical density)
This means that light travels 2.42 times slower in diamond than in vacuum.
Numerically, speed of light in diamond:
v = c / n
v = (3 × 10^8) / 2.42 ≈ 1.24 × 10^8 m/s
14. Find the power of a concave lens of focal length 2 m.
Answer:
For a concave lens, the focal length is negative:
f = -2 m
The power of the lens is given by:
P = 1 / f(m)
P = 1 / (-2) = -0.5 D
focal length) lens can produce an enlarged virtual image for near vision (magnifying effect).
Light Reflection and Refraction Class 10 SEBA Textual Exercise Solutions
Exercise
1. Which one of the following materials cannot be used to make a lens?
(a) Water
(b) Glass
(c) Plastic
(d) Clay
Answer: (d) Clay.
Reason: A lens must be transparent so light can pass through and be refracted. Clay is opaque.
2. 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.
Answer: (d) Between the pole of the mirror and its principal focus.
Reason: When an object is placed between pole and focus of a concave mirror, the mirror produces a virtual, erect and enlarged image.
3. Where should an object be placed in front of a convex lens to get a real image of the size of the object?
(a) At the principal focus of the lens
(b) At twice the focal length
(c) At infinity
(d) Between the optical centre of the lens and its principal focus.
Answer: (b) At twice the focal length.
Reason: For a convex (converging) lens, if object is at 2f (twice focal length) the real image is formed at 2f and is same size as object.
4. A spherical mirror and a thin spherical lens have each a focal length of –15 cm. The mirror and the lens are likely to be
(a) both concave.
(b) both convex.
(c) the mirror is concave and the lens is convex.
(d) the mirror is convex, but the lens is concave.
Answer: (d) the mirror is convex, but the lens is concave.
Reason: Negative focal length means the device is diverging: a convex mirror (diverging) and a concave lens (diverging).
5. No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be
(a) plane.
(b) concave.
(c) convex.
(d) either plane or convex.
Answer: (d) either plane or convex.
Reason: A plane mirror always gives an erect virtual image. A convex mirror also always gives an erect virtual image (although reduced). A concave mirror gives erect image only when object is between pole and focus.
SEBA Class 10 Science Chapter 10 Textbook Answers
6. Which of the following lenses would you prefer to use while reading small letters found in a dictionary?
(a) A convex lens of focal length 50 cm.
(b) A concave lens of focal length 50 cm.
(c) A convex lens of focal length 5 cm.
(d) A concave lens of focal length 5 cm.
Answer: (c) A convex lens of focal length 5 cm.
Reason: A strong convex (short focal length) lens can produce an enlarged virtual image for near vision (magnifying effect).
7. We wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the range of distance of the object from the mirror? What is the nature of the image? Is the image larger or smaller than the object? Draw a ray diagram to show the image formation in this case.
Answer: The object should be placed between the pole (P) and the focus (F) of the concave mirror.
Range of object distance:
The object distance should be less than 15 cm (0 to 15 cm).
Nature of image:
The image formed is virtual and erect.
Size of image:
The image is enlarged (larger than the object).
Ray diagram (how to draw):
Draw a concave mirror with principal axis and mark focus (F) at 15 cm.
Place the object between P and F.
Draw rays:
- A ray parallel to the principal axis reflects through F
- A ray towards centre of curvature reflects back on itself
- A ray through F reflects parallel to the principal axis
Extend reflected rays backward to meet behind the mirror to form a virtual, erect, enlarged image.
8. Name the type of mirror used in the following situations. Support your answer with reason.
(a) Headlights of a car.
(b) Side/rear-view mirror of a vehicle.
(c) Solar furnace.
Answer:
(a) Headlights of a car: Concave mirror
Reason: It produces a strong, parallel beam of light.
(b) Side/rear-view mirror: Convex mirror
Reason: It gives a wider field of view and forms erect, small images.
(c) Solar furnace: Concave mirror
Reason: It focuses sunlight at a point and produces high heat.
9. One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.
Answer: Yes, the lens will still produce a complete image.
Experimental observation:
When half of the convex lens is covered, the image is still complete but becomes less bright.
Explanation:
Every part of a lens forms the full image. Covering half only reduces the amount of light entering the lens, so the brightness decreases but the image remains complete.
10. An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.
Answer & calculation:
Lens formula:
1/f = 1/v + 1/u
Given:
Object distance, u = 25 cm
Focal length, f = 10 cm
Finding image distance, v:
1/v = 1/f – 1/u
1/v = 1/10 – 1/25
1/v = 0.1 – 0.04 = 0.06
v = 1 / 0.06 ≈ 16.666 cm (16 2/3 cm)
Magnification, m = v / u
m = 16.666 / 25 ≈ 0.6667
If the object size = 5 cm, then image size = m × 5 ≈ 3.333 cm
Position: 16.67 cm on the other side of lens (real image).
Nature: Real, inverted, and diminished (image smaller than object).
Ray diagram (description): draw principal axis, lens, object at 25 cm left, draw ray parallel→ through focus, ray through center→straight, and ray through focus→parallel; their intersection gives image at ~16.67 cm on right, inverted and smaller.
11. A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens? Draw the ray diagram.
Answer & calculation:
For a concave lens:
Focal length, f = -15 cm (negative)
The image formed by a concave lens is virtual, so image distance v is negative:
v = -10 cm
Using the lens formula:
1/f = 1/v – 1/u
Substitute the values:
1/(-15) = 1/(-10) + 1/u
1/u = -1/15 + 1/10
1/u = (-2 + 3)/30
1/u = 1/30
So, object distance:
u = 30 cm
Object distance: 30 cm in front of lens.
Nature of image: Virtual, erect, diminished, located 10 cm on the same side as the object.
Ray-diagram (description): For concave lens, draw parallel ray → refracts outward appearing from focus; ray aimed at centre goes straight. Extensions meet at virtual image position at 10 cm, upright and smaller.
12. An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.
Answer & calculation:
For a convex mirror:
Focal length, f = -15 cm (negative)
Using the mirror formula:
1/v + 1/u = 1/f
Given:
Object distance, u = 10 cm (object in front of mirror)
Find image distance, v:
1/v = 1/f – 1/u
1/v = -1/15 – 1/10
1/v = -1/6
v = -6 cm
- Position: image is 6 cm behind the mirror (virtual).
- Nature: Virtual, erect and diminished.
- Magnification: Linear magnification, m = -v / u
- m = -(-6) / 10
- m = 0.6
- So, image size = 0.6 × object size
HSLC Science Light Reflection Refraction Question Answer
13. The magnification produced by a plane mirror is +1. What does this mean?
Answer: Magnification +1 means the image formed is equal in size to the object.
The positive sign shows that the image is erect.
In a plane mirror, the image is virtual, erect, and of the same size as the object.
The image is formed at the same distance behind the mirror as the object is in front.
14. An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.
Answer & calculation:
Given:
Radius of curvature, R = 30 cm
Focal length, f = R/2 = 15 cm
(For convex mirror, f = -15 cm)
Object distance, u = 20 cm
Using the mirror formula:
1/v = 1/f – 1/u
1/v = -1/15 – 1/20
1/v = -7/60
v = -60/7 ≈ -8.571 cm
Position: image 8.571 cm behind mirror (virtual).
Nature: Virtual, erect, diminished.
Magnification:
m = -v / u
m = -(-60/7) / 20
m = 60 / 140
m = 3/7 ≈ 0.4286
Image size:
= 5.0 × 0.4286
≈ 2.14 cm
15. An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed, so that a sharp focused image can be obtained? Find the size and the nature of the image.
Answer & calculation:
Concave mirror:
f = +18 cm
u = 27 cm
Using the mirror formula:
1/v = 1/f – 1/u
1/v = 1/18 – 1/27
1/v = (3 – 2) / 54
1/v = 1/54
Therefore:
v = 54 cm
Magnification:
m = v / u
m = 54 / 27 = 2
Image size:
= m × 7.0
= 2 × 7.0 = 14.0 cm
Screen distance: 54 cm in front of mirror (to catch the real image).
Nature: Real, inverted, and magnified (image twice the object size).
16. Find the focal length of a lens of power – 2.0 D. What type of lens is this?
Answer:
Power:
P = -2.0 D
Focal length:
f = 1 / P
f = 1 / (-2.0)
f = -0.5 m = -50 cm
Type:
Negative power ⇒ concave (diverging) lens
17. A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?
Answer:
Focal length:
f = 1 / P
f = 1 / 1.5
f = 0.6667 m ≈ 66.7 cm
Positive power ⇒ converging (convex) lens
🔬 Updated Solutions Notice: This page features complete, step-by-step SEBA Class 10 Science Chapter 10 (Light – Reflection and Refraction) Textual Exercise Solutions updated for the 2026–27 academic session. All textbook question answers, numerical calculations, and ray diagrams strictly follow the latest revised Assam Board (SEBA) curriculum.
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