# NCERT Solutions for Class 8 Science Chapter 16 Light

The NCERT Solutions for Class 8 Science Chapter 16 Light is extremely crucial for the students studying in CBSE Class 8. The Chapter 16 (Light) Solutions are provided here to help the students to clear all their doubts and to help them in understanding the chapter in an easy and interesting way.

While solving the textbook questions, students often face several doubts and eventually get confused. To help the students clear all their doubts instantly, the NCERT Solutions come as a great resource. Here, the detailed solutions for the questions in Chapter 16 ‘Light’ from NCERT Science book Class 8 are given.

These NCERT Solutions present you with answers to the question on the meaning of reflection, laws of reflection, types of mirrors, kaleidoscope and its construction, Concept of the rainbow and 7 colours, Dispersion of light, Parts of human eye and defects of the human eye.

### Access Answers to NCERT Class 8 Science Chapter 16 Light

Exercise Questions

1. Suppose you are in a dark room. Can you see objects in the room? Can you see objects outside the room? Explain.

Soln:

If a person is inside the room where there is no light, it is then impossible to visualize the object inside the room but the object out of the room can be seen easily.

When light falls on eyes after reflecting from the object, it becomes visible. If the room is dark, then the object which is in the room reflects no light. Hence, the person is not able to see the objects in the room where there is no light.

2. Differentiate between regular and diffused reflection. Does diffused reflection mean the failure of the laws of reflection?

Soln:

 S.No Regular Reflection Diffused Reflection 1. It occurs when the surface is smooth. It occurs when the surface is a rough surface. 2. Reflected rays move in a particular direction. Reflected rays scattered in random directions. Example: Reflection by the plane mirror Example: Reflection by the road surface.

The laws of reflection have not failed because each ray obeys the law of reflection. All the reflected rays are parallel to each other, whereas in diffused reflection the rays aren’t parallel to the incident rays.

3. Mention against each of the following whether regular or diffused reflection will take place when a beam of light strikes. Justify your answer in each case.

(a) Polished wooden table

(b) Chalk powder

(c) Cardboard surface

(d) Marble floor with water spread over it

(e) Mirror

(f) Piece of paper

Soln:

a) The wooden table that has been a polished-Regular reflection

The surface that has been recently polished can be a good example of a smooth surface. The wooden table that has been polished has a surface that is smooth.

b) White Chalk powder that is used in school- Diffused reflection

Chalk powder spread on a surface is an example of an irregular surface. Hence, it is rough. Therefore, the diffused reflection will appear from chalk powder.

c) Cardboard surface- Diffused reflection

The surface of the cardboard is a kind of irregular surface. Hence, the diffused reflection will take place from a cardboard surface.

d) Marble floor – Regular reflection

Marble floor can be a good example of a surface that is regular. Since water makes the ceramic glossy, thus, the reflections that are regular occur on this surface.

e) Mirror- Regular reflection

A mirror has a very smooth surface hence it gives a regular reflection.

f) Piece of paper- Diffused reflection

Although a piece of paper may look smooth, it has many irregularities on its surface. Due to this reason, it will give a diffused reflection.

4. State the laws of reflection.

Soln:

The law of reflection states that

a) The angle of reflection and the angle of incidence both are always equal to one another.

b) The reflected ray, the incident ray, and the normal to the reflective surface at the point of incidence all come on the same plane.

5. Describe an activity to show that the incident ray, the reflected ray and the normal at the point of incidence lie in the same plane.

Soln:

On a table, place a plane mirror perpendicular to the plane of the table. Make a small hole in a paper and hold it perpendicular to the plane of the table. Try to do this experiment in a dark room. Take one more piece of paper and place it on the table so that it makes contact with the mirror. Draw a line perpendicular to the mirror on the piece of paper which is on the table. Now beam light rays with the help of a torch through the small hole such that the beam of light hits the normal at the bottom of the mirror. The ray of light will be reflected in the light rays from the hole are incident on the mirror. Looking at the piece of paper on the table, we can easily show that the incident ray, the normal line and the reflected ray at the point of incidence lie in the same plane.

6. Fill in the blanks in the following.

(a) A person 1 m in front of a plane mirror seems to be _______________ m away from his image.

(b) If you touch your ____________ ear with right hand in front of a plane mirror it will be seen in the mirror that your right ear is touched with ____________.

(c) The size of the pupil becomes ____________ when you see in dim light.

(d) Night birds have ____________ cones than rods in their eyes.

Soln:

(a) A person 1 m in front of a plane mirror seems to be 2m away from his image.

(b) If you touch your left ear with right hand in front of a plane mirror it will be seen in the mirror that your right ear is touched with the left hand.

(c) The size of the pupil becomes large when you see in dim light.

(d) Night birds have fewer cones than rods in their eyes.

Choose the correct option in Questions 7 – 8

7. The angle of incidence is equal to the angle of reflection.

(a) Always

(b) Sometimes

(c) Under special conditions

(d) Never

Soln:

Answer is (a) Always

8. Image formed by a plane mirror is

(a) virtual, behind the mirror and enlarged.

(b) virtual, behind the mirror and of the same size as the object.

(c) real at the surface of the mirror and enlarged.

(d) real, behind the mirror and of the same size as the object.

Soln:

Answer is (b) virtual, behind the mirror and of the same size as the object.

9. Describe the construction of a kaleidoscope.

Soln:

The construction of a kaleidoscope:

Take three rectangular mirror strips of dimensions 15cm x 4cm (l x b) and join them together to form a prism. A prism is fixed into a circular cardboard tube. The circular cardboard tube should be slightly longer than the prism. This circular tube is now closed at one end with a cardboard disc. This disc has a hole through which we can see. At the other end of the circular tube, a plane glass plate is fixed. It is important that this glass plate touches the prism mirrors. On this glass plate, several small and broken pieces of coloured glass are placed. This end is now closed by a round glass plate allowing enough space for the coloured glass pieces to move.

10. Draw a labelled sketch of the human eye.

Soln:

11. Gurmit wanted to perform Activity 16.8 using a laser torch. Her teacher advised her not to do so. Can you explain the basis of the teacher’s advise?

Soln:

Her teacher advised her not to do so because of the intensity of the laser light is very high, it is harmful to the human eyes. It can cause damage to the retina and leads to blindness. Hence, it is advisable not to look at a laser beam directly.

12. Explain how you can take care of your eyes.

Soln:

The following points help a person to take care of his eyes:

a) Reading should not be done in bright light as well as in dim light.

b) He should visit an eye specialist on a regular interval of time.

c) If any small insects or dust particles enters his eyes, do not rub them but clean them immediately with cold water.

d) He should avoid direct exposure of sunlight to the eye.

e) While reading, there should be a distance of at least 25 cm between the eyes and the book.

13. What is the angle of incidence of a ray if the reflected ray is at an angle of 90° to the incident ray?

Soln:

If the reflected ray is at the angle of 90to the incident ray, then the angle of incidence is 45o. According to the law of reflection, the angle of incidence and the angle of reflection are equal. Therefore, the angle of incidence and the angle of reflection both are 90/2=45o.

14. How many images of a candle will be formed if it is placed between two parallel plane mirrors separated by 40 cm?

Soln:

If a candle is placed between two parallel plane mirror separated by 40 cm, then the multiple and infinite images will be formed due to the multiple reflections between the mirrors. The infinite numbers of images are formed when two mirrors are placed parallel to each other.

15. Two mirrors meet at right angles. A ray of light is incident on one at an angle of 30° as shown in Fig. 16.19. Draw the reflected ray from the second mirror.

Soln:

The first law of reflection is used to obtain the path of reflected light.

It can be observed that the given ray of light will reflect from the second mirror at an angle of 60°.

The incident ray OA reflects at point O
Since Angle of Reflection = Angle of Incidence
OO’ makes an angle 300 with normal of first mirror
Now, drawing normal at O’
The two normals intersect at 900 angle
And applying angle sum property in △ OXO’
We get
Angle of Incidence in 2nd mirror = 600
Applying law of reflection in 2nd mirror,
We get the below figure

16. Boojho stands at A just on the side of a plane mirror as shown in Fig. 16.20. Can he see himself in the mirror? Also, can he see the image of objects situated at P, Q and R?

Soln:

Boojho cannot see his image because the reflected ray won’t reach his eyes.He can see the image of objects situated at P, Q because the rays coming from P and Q get reflected by the mirror and reach his eyes. Boojho can’t see the image of objects situated at R because the ray from object R does not get reflected.

17. (a) Find out the position of the image of an object situated at A in the plane mirror (Fig. 16.21).

(b) Can Paheli at B see this image?

(c) Can Boojho at C see this image?

(d) When Paheli moves from B to C, where does the image of A move?

Soln:

a) Image of an object placed at A is formed behind the mirror. The distance of the image from the mirror is equal to the distance of A from the mirror

b)Yes Paheli at B can see this image.

c) Yes Boojho at C can see this image.

d) Image of the object at A will not move. It will remain at the same position when Paheli moves from B to C.

 Also Access NCERT Exemplar for class 8 Science Chapter 16 CBSE Notes for class 8 Science Chapter 16

### Class 8 Science NCERT Solutions for Chapter 16 Light

These solutions are created by subject experts according to the latest CBSE syllabus. This solution consists of answers to the textbook questions, extra questions, exemplary problems, worksheets and questions from previous year question papers.

### Subtopics of NCERT Class 8 Science Chapter 16 Light

1. What makes Things Visible
2. Laws of Reflection
3. Regular and Diffused Reflection
4. Reflected Light can be Reflected Again
5. Multiple Images
6. Sunlight – White or Coloured
7. What is inside Our Eyes
8. Care of the Eyes
9. Visually Challenged Persons can Read and Write
10. What is the Braille System?

Chapter 16 Light is one of the most important chapters in the CBSE Class 8 Science book. Several important topics related to light are introduced to the students, which are not only important from the examination point of view but also to understand the concepts in higher classes.

First, students are introduced to the concept of light and then to the laws of reflection. This chapter introduces  students to the incident rays, the angle of incidence and reflection, lateral inversion, diffused (irregular)and regular reflection.

Apart from these, students also get to know about the human eye and its different parts like pupil, cornea, iris, retina, etc. The Class 8 students are also taught how to take care of the eyes and introduced to the braille system. The different activities included in the curriculum are important as they help the students to get engaged with the learning to have in-depth knowledge about different concepts. It is important for the students to solve different questions given in the book and refer to these solutions in case of any doubts. Stay tuned with CoolGyan’S to get the complete NCERT Solutions to get additional assistance for the exams. Students are also advised to download CoolGyan’S- The Learning App to have an effective and personalized learning experience.

## Frequently Asked Questions on NCERT Solutions for Class 8 Science Chapter 16

### What are the benefits of using the NCERT Solutions for Class 8 Science Chapter 16?

The benefits of using the NCERT Solutions for Class 8 Science Chapter 16 are –
1. Solutions created in easy to learn format.
2. The subject matter experts have crafted the solutions as per the CBSE syllabus.
3. MCQs have been answered with proper reasoning.
4. Previous year question papers and sample papers are also available for the respective subject.

### What are the topics covered under NCERT Solutions for Class 8 Science Chapter 16?

The topics covered under NCERT Solutions for Class 8 Science Chapter 16 are –
What makes Things Visible
Laws of Reflection
Regular and Diffused Reflection
Reflected Light can be Reflected Again
Multiple Images
Sunlight – White or Coloured
What is inside Our Eyes
Care of the Eyes
Visually Challenged Persons can Read and Write
What is the Braille System?

### Explain law of reflection in Chapter 16 of NCERT Solutions for Class 8 Science?

The principle when the light rays falls on the smooth surface, the angle of reflection is equal to the angle of incidence, also the incident ray, the reflected ray, and the normal to the surface all lie in the same plane. The law of reflection defines that upon reflection from a smooth surface, the angle of the reflected ray is equal to the angle of the incident ray, with respect to the normal to the surface that is to a line perpendicular to the surface at the point of contact.
The reflected ray is always in the plane defined by the incident ray and normal to the surface at the point of contact of the incident ray.

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