The magnification produced when object is placed at a distance of 20 cm from a spherical mirror is + 1 2 . Where should the object be placed to reduce the magnification to + 1 3 . When an object is placed at a distance of 50 cm from a concave spherical mirror, the magnification produced is - 1 2 . Where should the object be placed to get a magnification of - 1 5 ?

Given: Object distance, u = 20 cm Magnification, m = + 1 2 Let the image distance is v and the focal length is f . The magnification, m = − v u That is, + 1 2 = − v u = − v 20 1 2 = − v 20 v = - 20 2 = - 10 cm Substituting this value of v into the magnification equation, we get, + 1 3 = - ( - 10 ) u 1 3 = 10 u Simplifying, we get: u = 30 cm Therefore, the object should be placed at a distance of 30 cm from the spherical mirror to reduce the magnification to + 1 3 .

Draw the path of a ray of light when it enters one of the faces of a glass slab at an angle of nearly 45 ° . Label on it (i) angle of refraction, (ii) angle of emergence and (iii) lateral displacement.

The required diagram is given below: The change in the direction of light when it passes from one medium to another is known as the refraction of light. The angle between incident ray and normal is called the angle of incidence, represented as i . Here i = 45 ° . The angle between refracted ray and normal is called the angle of refraction, represented as r . The angle of the light coming out of a medium is called the angle of emergence. The perpendicular distance of separation between the emergent ray and the incident ray is known as lateral displacement.

If the image formed by a spherical mirror for all positions of the object placed in front of it is always erect and diminished, what type of mirror is it ? Draw a labelled ray diagram to support your answer.

Convex mirror always forms erect and diminished image. In the above ray diagram, the image formed is behind the mirror between pole and focus, virtual, erect and diminished.

What is meant by power of a lens? Write its SI unit. A student uses a lens of focal length 40 cm and another of - 20 cm . Write the nature and power of each lens.

The power of a lens is a measure of the degree of convergence or divergence of light rays falling on it. It is also defined as the reciprocal of its focal length in metres. The S.I. unit of power is dioptre. Focal length of a convex lens is positive. F c o n v e x = 40 cm = 40 100 m Power = 1 focal length = 1 ( 40 100 ) Power = + 2 . 5 D So, lens having focal length 40 cm is convex. Focal length of a concave lens is negative. F c o n c a v e = - 20 cm = - 20 100 m Power = 1 focal length = 1 ( - 20 100 ) Power = - 5 D Converging (convex ) lenses have positive focal lengths, so they also have positive power values. Diverging (concave ) lenses have negative focal lengths, so they also have negative power values.

State the laws of refraction of light. Explain the term 'absolute refractive index of a medium' and write an expression to relate it with the speed of light in vacuum.

The two laws of refraction are: The Incident ray, the refracted ray and the normal at the point of incidence, all lie in the same plane. The ratio of the sine of angle of incidence( i ) to the sine of angle of refraction ( r ) at a point in a medium is constant, i.e., sin i sin r = C o n s t a n t . Absolute refractive index is the ratio of speed or velocity of light in vacuum to the speed of light in the given medium. A b s o l u t e r e f r a c t i v e i n d e x ( η ) = S p e e d o f l i g h t i n v a c u u m o r a i r ( c ) S p e e d o f l i g h t i n m e d i u m ( v ) .

A 10 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 12 cm . The distance of the object from the lens is 18 cm . Find the nature, position and size of the image formed.

Given, The height of the object, h 1 = + 10 cm Focal length, f = + 12 cm Object distance, u = - 18 cm We have the lens formula, 1 v - 1 u = 1 f ⇒ 1 v - 1 - 18 = 1 12 ⇒ 1 v + 1 18 = 1 12 ⇒ 1 v = 1 12 - 1 18 ⇒ 1 v = 3 - 2 36 = 1 36 ∴ v = 36 cm Magnification, m = v u = 36 - 18 = - 2 Thus, we can say that the position of the image is formed at a distance of 36 cm from the convex lens. As the value of the magnification is more than 1 , the image formed is larger than the object. The minus sign of magnification indicates that the image is formed below the principal axis. Thus, we can say that the image formed is real as it can be captured on the screen, and it is inverted.

A student holding a mirror in his hand, directed the reflecting surface of the mirror towards the Sun. He then directed the reflected light on to a sheet of paper held close to the mirror. Will he be able to determine the approximate value of focal length of this mirror from this activity ? Give reason and draw ray diagram to justify your answer in this case.

Yes, the sheet of paper will start burning only when he kept it at the focus of the mirror. This will give an approximate value of focal length which is the distance between the mirror and the point where the paper starts burning. Here, we are using a concave mirror because it is having converging properties. The required ray diagram is given below: A real image of the sun is formed by the concave mirror.

A student holding a mirror in his hand, directed the reflecting surface of the mirror towards the Sun. He then directed the reflected light on to a sheet of paper held close to the mirror. Which type of mirror does he have?

The student is having a concave mirror in his hand. Concave mirror is having converging properties. Hence, in order to burn the paper, the student can place the paper at the principal focus of the concave mirror where the intensity of light will be maximum.

A student holding a mirror in his hand, directed the reflecting surface of the mirror towards the Sun. He then directed the reflected light on to a sheet of paper held close to the mirror. What should he do to burn the paper?

Mirror that is having the converging properties will be used to burn the paper. This is because to burn the paper we need to converge the intensity of sunlight at one point. This type of mirror is a concave mirror. The student has to place the paper at such a point where the intensity of light will be maximum, that is the principal focus of the mirror.

HARD

10th CBSE

IMPORTANT

Earn 100

An object of height $10 cm$ is placed $25 cm$ away from the optical centre of a converging lens of focal length $15 cm$ . Calculate the image-distance and height of the image formed.

Important Questions on Light - Reflection and Refraction

HARD

10th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR SCIENCE>Chapter 10 - Light - Reflection and Refraction>CBSE-2023 (Delhi - I)>Q 31

The power of a lens is

+ 4 D

. Find the focal length of this lens. An object is placed at a distance of

50 cm

from the optical centre of this lens. State the nature and magnification of the image formed by the lens and also draw a ray diagram to justify your answer.

HARD

10th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR SCIENCE>Chapter 10 - Light - Reflection and Refraction>CBSE-2023 (Delhi - II)>Q 35

List two difference in the characteristic properties of the virtual images formed by the two types of spherical lenses (concave and convex). How are these characteristics of the two lenses use in the correction of the two common defects of vision namely myopia and hypermetropia?

HARD

10th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR SCIENCE>Chapter 10 - Light - Reflection and Refraction>CBSE-2023 (Delhi - II)>Q 37

Define the following terms in the context of a diverging lens:

(i) Principal focus.

(ii) Focal length.

Draw a labelled ray diagram to illustrate your answer.

HARD

10th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR SCIENCE>Chapter 10 - Light - Reflection and Refraction>CBSE-2023 (Delhi - III)>Q 38

The magnification produced when object is placed at a distance of $20 cm$ from a spherical mirror is $+ \frac{1}{2}$ . Where should the object be placed to reduce the magnification to $+ \frac{1}{3}$ .

When an object is placed at a distance of $50 cm$ from a concave spherical mirror, the magnification produced is $- \frac{1}{2}$ . Where should the object be placed to get a magnification of $- \frac{1}{5}$ ?

MEDIUM

10th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR SCIENCE>Chapter 10 - Light - Reflection and Refraction>CBSE-2023 (Outside Delhi - I)>Q 39

A student wants to obtain an erect image of an object using a concave mirror of

10 cm

focal length. What will be the distance of the object from mirror?

HARD

10th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR SCIENCE>Chapter 10 - Light - Reflection and Refraction>CBSE-2023 (Outside Delhi - I)>Q 41

A ray of light starting from diamond is incident on the interface separating diamond and water. Draw a labelled ray diagram to show the refraction of light in this case.

HARD

10th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR SCIENCE>Chapter 10 - Light - Reflection and Refraction>CBSE-2023 (Outside Delhi - I)>Q 42

Absolute refractive indices of diamond and water are

2.42

and

1.33

respectively. Find the value of refractive index of water w.r.t. diamond.

EASY

10th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR SCIENCE>Chapter 10 - Light - Reflection and Refraction>CBSE-2023 (Outside Delhi - I)>Q 44

In human eye the part which allows light to enter into the eye is:

An object of height 10 cm is placed 25 cm away from the optical centre of a converging lens of focal length 15 cm. Calculate the image-distance and height of the image formed.

Important Questions on Light - Reflection and Refraction

Learn and Prepare for any exam you want !

An object of height $10 cm$ is placed $25 cm$ away from the optical centre of a converging lens of focal length $15 cm$ . Calculate the image-distance and height of the image formed.