This illuminating chapter of NCERT Solutions for Class 10 Science Chapter 9, Light – Reflection and Refraction, helps us understand the fascinating behavior of light as it interacts with different surfaces and materials. Whether it’s seeing your reflection in a mirror, understanding how spectacles correct vision, explaining why a pencil appears bent in water, or how lenses in cameras capture images, this chapter unveils the principles behind everyday optical phenomena. This chapter is part of the comprehensive NCERT Solutions Class 1o Science series, which covers all chapters in detail.
The chapter equips students with essential skills to apply laws of reflection, understand image formation by plane and spherical mirrors, work with mirror and lens formulas, comprehend refraction through different media, and solve numerical problems involving focal length, object distance, and image characteristics. Every solution has been designed keeping CBSE board exam patterns in mind, ensuring students develop both conceptual clarity and problem-solving confidence. By mastering this chapter, students build a strong foundation for optics, wave theory, and advanced physics topics in higher classes.
NCERT Solutions for Class 10 Science Chapter 9 - All Exercise Questions
Q.
Which one of the following materials cannot be used to make a lens?
(a) Water
(b) Glass
(c) Plastic
(d) Clay
Q.
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.
Q.
Find the focal length of a lens of power – 2.0 D. What type of lens is this?
Q.
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 focussed image can be obtained? Find the size and the nature of the image.
Q.
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.
Q.
The magnification produced by a plane mirror is +1. What does this mean?
Q.
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.
Q.
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.
Q.
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.
Q.
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 principle 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 principle focus.
Q.
Name the type of mirror used in the following situations.
(a) Headlights of a car
(b) Side/rear-view mirror of a vehicle.
(c) Solar furnace
Support your answer with reason.
Q.
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.
Q.
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 of 50 cm
(c) A convex lens of focal length of 5 cm
(d) A concave lens of focal length 5 cm
Q.
No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be-
(a) only plane.
(b) only concave.
(c) only convex.
(d) either plane or convex.
Q.
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 and the lens is concave.
Q.
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 principle focus of the lens
(b) At twice the focal length
(c) At infinity
(d) Between the optical centre of the lens and its principle focus.
Q.
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?
Class 10 Chapter 9 Science Questions & Answers –Light - Reflection and Refraction
Q1.Which one of the following materials cannot be used to make a lens?
(a) Water
(b) Glass
(c) Plastic
(d) Clay
Solution:The correct option is (d).
Explanation: The light can pass through a lens. As clay does not allow light to pass through it hence, it cannot be used to make a lens.
Q2. 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.
Solution: Given:
Object distance, \( u = -25\,\text{cm} \)
Object height, \( h_o = 5\,\text{cm} \)
Focal length, \( f = +10\,\text{cm} \)
Using the lens formula:
\[
\frac{1}{v} - \frac{1}{u} = \frac{1}{f}
\]
\[
\frac{1}{v} = \frac{1}{f} + \frac{1}{u}
\]
\[
\frac{1}{v} = \frac{1}{10} + \frac{1}{-25}
\]
\[
\frac{1}{v} = \frac{1}{10} - \frac{1}{25}
\]
\[
\frac{1}{v} = \frac{5 - 2}{50} = \frac{3}{50}
\]
\[
v = \frac{50}{3} \approx 16.66\,\text{cm}
\]
Since image distance \(v\) is positive, the image is formed on the other side of the lens.
Magnification:
\[
m = \frac{v}{u} = \frac{16.66}{-25} \approx -0.66
\]
The negative sign of magnification shows that the image is real and inverted.
Again,
\[
m = \frac{h_i}{h_o}
\]
\[
h_i = m \times h_o = -0.66 \times 5 \approx -3.3\,\text{cm}
\]
The negative sign of \(h_i\) shows that the image is inverted.
Hence, the image is formed:
- at \(16.66\,\text{cm}\) on the other side of the lens
- size \(\approx 3.3\,\text{cm}\)
- nature: real, inverted and diminished.
Q3. Find the focal length of a lens of power –2.0 D. What type of lens is this?
Solution:
Given, power of lens:
\[
P = -2\,\text{D}
\]
Using the formula:
\[
P = \frac{1}{f(\text{metres})}
\]
So,
\[
f = \frac{1}{P} = \frac{1}{-2} = -0.5\,\text{m}
\]
A concave lens always has a negative focal length.
Hence, this lens is concave.
Q4.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 nature of the image.
Solution
Given:
Object distance, \(u = -27\,\text{cm}\)
Object height, \(h_o = 7\,\text{cm}\)
Focal length, \(f = -18\,\text{cm}\)
Using the mirror formula:
\[
\frac{1}{v} + \frac{1}{u} = \frac{1}{f}
\]
\[
\frac{1}{v} = \frac{1}{f} - \frac{1}{u}
\]
\[
\frac{1}{v} = \frac{1}{-18} - \frac{1}{-27}
\]
\[
\frac{1}{v} = -\frac{1}{18} + \frac{1}{27}
\]
\[
\frac{1}{v} = -\frac{1}{54}
\]
\[
v = -54\,\text{cm}
\]
Hence, the screen should be placed 54 cm in front of the mirror.
Magnification:
\[
m = -\frac{v}{u} = -\left(\frac{-54}{-27}\right) = -2
\]
The negative sign of magnification shows that the image is real.
Again,
\[
m = \frac{h_i}{h_o}
\]
\[
h_i = m \times h_o = -2 \times 7 = -14\,\text{cm}
\]
The negative value of image height indicates that the image is inverted.
Final Results:
- Image distance: \(54\,\text{cm}\) in front of the mirror
- Image height: \(14\,\text{cm}\) (inverted)
- Nature of image: Real, inverted, enlarged
Q5. 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.
Solution:
Given:
Object distance, \(u = -20\,\text{cm}\)
Object height, \(h_o = 5\,\text{cm}\)
Radius of curvature, \(R = 30\,\text{cm}\)
Focal length:
\[
f = \frac{R}{2} = \frac{30}{2} = 15\,\text{cm}
\]
Using the mirror formula:
\[
\frac{1}{v} + \frac{1}{u} = \frac{1}{f}
\]
\[
\frac{1}{v} = \frac{1}{f} - \frac{1}{u}
\]
\[
\frac{1}{v} = \frac{1}{15} - \frac{1}{-20}
\]
\[
\frac{1}{v} = \frac{1}{15} + \frac{1}{20}
\]
\[
\frac{1}{v} = \frac{7}{60}
\]
\[
v = 8.57\,\text{cm}
\]
Since \(v\) is positive, the image is formed behind the mirror.
Magnification:
\[
m = -\frac{v}{u} = -\left(\frac{8.57}{-20}\right) = 0.428
\]
Positive magnification ⇒ image is virtual.
Again,
\[
m = \frac{h_i}{h_o}
\]
\[
h_i = m \times h_o = 0.428 \times 5 = 2.14\,\text{cm}
\]
The positive value of \(h_i\) shows the image is **erect**.
Therefore, the image formed is:
- virtual
- erect
- smaller in size (diminished)
- located 8.57 cm behind the mirror
Q6.The magnification produced by a plane mirror is +1. What does this mean?
Solution:
For a mirror, magnification is given by:
\[
m = \frac{h_i}{h_o}
\]
where
\(h_i\) = image height
\(h_o\) = object height
If the magnification produced by a plane mirror is:
\[
m = +1
\]
This means:
- The image formed is of the same size as the object, because \(h_i = h_o\)
- The positive sign indicates that the image is virtual and erect
- This is a characteristic property of plane mirrors
Q7. 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.
Solution:
Given:
Focal length, \(f = +15\,\text{cm}\)
Object distance, \(u = -10\,\text{cm}\)
Using the mirror formula:
\[
\frac{1}{v} + \frac{1}{u} = \frac{1}{f}
\]
\[
\frac{1}{v} = \frac{1}{f} - \frac{1}{u}
\]
\[
\frac{1}{v} = \frac{1}{15} - \frac{1}{-10}
\]
\[
\frac{1}{v} = \frac{1}{15} + \frac{1}{10}
\]
\[
\frac{1}{v} = \frac{2 + 3}{30} = \frac{5}{30}
\]
\[
v = 6\,\text{cm}
\]
Since \(v\) is positive, the image is formed behind the mirror.
Magnification:
\[
m = -\frac{v}{u} = -\left(\frac{6}{-10}\right) = +0.6
\]
The positive sign of magnification shows that the image is virtual and erect.
Final nature of the image:
- Virtual
- Erect
- Diminished
- Formed 6 cm behind the mirror
Q8. 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.
Solution:
Given:
Focal length of concave lens, \(f = -15\,\text{cm}\)
Image distance, \(v = -10\,\text{cm}\)
Using the lens formula:
\[
\frac{1}{v} - \frac{1}{u} = \frac{1}{f}
\]
Substituting values:
\[
\frac{1}{-10} - \frac{1}{u} = \frac{1}{-15}
\]
\[
-\frac{1}{10} - \frac{1}{u} = -\frac{1}{15}
\]
Rearranging:
\[
-\frac{1}{u} = -\frac{1}{15} + \frac{1}{10}
\]
\[
-\frac{1}{u} = -\frac{2}{30} + \frac{3}{30}
\]
\[
-\frac{1}{u} = \frac{1}{30}
\]
\[
u = -30\,\text{cm}
\]
The negative object distance indicates that the object is placed 30 cm in front of the concave lens.
Q9. 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.
Solution:
The convex lens will produce a compete image of the object kept in front of it even half of the lens is covered with black paper.
i. When the upper half of the lens is covered:
The rays of light coming from the object will be refracted by the lower half of the lens. These rays meet at the other side of the lens to form the image of the given object.
ii. When the lower half of the lens is covered:
The rays of light coming from the object are refracted by the upper half of the lens. These rays meet at the other side of the lens to form the image of the given object.
Q10. 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 principle 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 principle focus.
The correct option is (d).
Explanation: Concave mirror forms a virtual, erect and enlarged image of an object when the object is placed between the pole and focus of the mirror.
Q11. Name the type of mirror used in the following situations.
(a) Headlights of a car
(b) Side/rear-view mirror of a vehicle.
(c) Solar furnace
Support your answer with reason.
Solution:
(a) Concave mirror
(b) Convex mirror
(c) Concave mirror.
Reasons:
(a) In the headlights of cars, concave mirror is used as it produces parallel beam of light when the light source (bulb) is at the principal focus of the concave mirror.
(b) A convex mirror forms a virtual, erect, and diminished image of the objects placed in front of. Hence, a driver can see most of the traffic behind him.
(c) A concave mirror can converge the light incident on it at a single point known as principal focus. Therefore, it can be used to produce a large amount of heat at that point.
Q12. 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.
Solution:
A concave mirror always forms an erect image when object is placed between pole (P) and the principal focus (F). Therefore, the object should be placed anywhere between the pole and the focus of the concave mirror. The image formed in this case will be virtual, erect, and magnified in nature.
Q13. 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 of 50 cm
(c) A convex lens of focal length of 5 cm
(d) A concave lens of focal length 5 cm
The correct option is (c).
Explanation: When a convex lens is placed between the radius of curvature and focal length, it forms magnified image of the given object. Moreover, magnification is more for convex lenses having shorter focal length.
Q14. No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be-
(a) only plane.
(b) only concave.
(c) only convex.
(d) either plane or convex.
The correct option is (d).
Explanation: A plane mirror always forms a virtual and erect image of same size as that of the object. In the same way, a convex mirror forms a virtual and erect image of smaller size of the object placed in front of it.
Q15. 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 and the lens is concave.
The correct option is (a).
Explanation: As per sign convention, the focal length of a concave mirror and a concave lens are considered as negative.
Q16. 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 principle focus of the lens
(b) At twice the focal length
(c) At infinity
(d) Between the optical centre of the lens and its principle focus.
The correct option is (b).
Explanation: On placing the object at a distance twice of the focal length (at centre of curvature) in front of a convex lens, the image is formed at the centre of curvature on the other side of the lens. This image is real, inverted, and of the same size as the object.
Q17. 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?
Solution:
Given:
Power of lens, \(P = +1.5\,\text{D}\)
Using the power formula:
\[
P = \frac{1}{f(\text{metres})}
\]
\[
f = \frac{1}{P} = \frac{1}{1.5} = 0.66\,\text{m}
\]
Since the focal length is positive, the lens is converging (convex lens).
More Resources of NCERT Solutions for Class 10 Science
NCERT Solutions for Class 10 Science Chapter 9 – FAQs
Q1. What is the difference between real and virtual images in mirrors and lenses?
A real image is formed when light rays actually converge at a point after reflection or refraction. It can be projected on a screen and is always inverted. Real images are formed by concave mirrors (when object is beyond focus) and convex lenses.
A virtual image is formed when light rays appear to diverge from a point but don't actually meet. It cannot be projected on a screen and is always erect. Virtual images are formed by plane mirrors, convex mirrors, concave lenses, and concave mirrors (when object is between pole and focus).
Q2. How do I use the mirror formula and lens formula to solve numerical problems?
The mirror formula is: 1/f = 1/v + 1/u
The lens formula is: 1/f = 1/v - 1/u
Where f = focal length, v = image distance, u = object distance
Key sign conventions to remember:
- Distances measured against the direction of incident light are negative
- Distances measured in the direction of incident light are positive
- Heights measured upward from principal axis are positive, downward are negative
Always apply sign conventions correctly before substituting values, and remember that focal length is negative for concave mirrors and positive for convex mirrors.
Q3. What are the practical applications of spherical mirrors and lenses in daily life?
Concave mirrors are used in torches, headlights, shaving mirrors, dentist mirrors, and solar concentrators because they can focus light or magnify images.
Convex mirrors are used as rear-view mirrors in vehicles and at road intersections because they provide a wider field of view and always form diminished, erect images.
Convex lenses are used in magnifying glasses, cameras, telescopes, microscopes, and to correct hypermetropia (far-sightedness), while concave lenses are used to correct myopia (near-sightedness).