Light: Refraction and Reflection || Exercise Question || CBSE/NCERT

Light is a form of energy which produced in us the sensation of vision. Speed of light in vacuum is 3 x 10^8 meter per second. White light of sunlight consist of seven colors.

Ques.1:_ Which one of the following materials cannot be used to make a lens?

(a) Water

(b) Glass

(d) Clay

Answer:_ A lens allow light to pass through it since clay does not show such properties; it cannot be used to make a lens.

Ques.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.

(d) Between the pole of the mirror and its principal focus.

Answer:_ It should be between the principal focus and centre of curvature.

Question 3:_ Where should an object be placed in the front of convex lens to get a real image of the size of the object?

(a) At the principal focus of lens.

(b)At twice the focal length.

(d) Between the optical centre of the lens and its principal axis.

Answer:_ (b) At twice the focal length i.e. at the centre of curvature.

Question 4:_ A spherical mirror and a thin spherical lens have each a focal length ─15 cm. The mirror and the lens are likely to be:-

(a) both concave

(b) Both convex

(d) the mirror is convex but the lens is concave.

Answer:_ (a) both concave. As focal length of the concave lens and a concave mirror are taken as negative.

Question 5:_ No matter how far stand from a mirror, your image appears erect. The mirror is likely to be:

(a) Plane

(b) Concave

(d) either plane or convex.

Answer: _ (d) either plane or convex.

Question 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.

(d) A concave lens of focal length 5 cm

Answer:_ A convex lens of focal length 5 cm would prefer to use while reading small letters found in dictionary.

Question 7:_ We wise to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the range of the 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:_ Range of object distance = 0 cm to 15 cm.

A concave mirror give an erect image when an object is placed between its pole (P) and the principal focus.

Hence, to obtain an erect image of an object from a concave mirror of focal length 15 cm, the object must be placed anywhere between the pole and the focus. The image formed will be virtual erect and magnified in nature as shown in figure.

Question 8:_ Name the type of mirror used in the following situation:

(a) Headlight of a car

(b) Side/rear-view mirror of a vehicles.

Answer:_ (a) Headlight of a car: Concave

Reason: Concave mirror is used in headlight of a car. This is because concave mirror can produced powerful parallel a beam of light. When the light source is placed at their principal focus.

(b) Side/rear-view mirror of vehicles: Convex

Reason: Convex mirror give a virtual, erect and diminished image of the object placed in front of it. Because of this, they have a wide field of view. It enables the driver to see most of the traffic behind him/her.

(c) Solar furnace: Concave

Reason: Concave mirror are convergent mirrors. That is why they are used to construct solar furnaces. Concave mirrors converge the light incident on them at a single point known as principal focus. Hence they can be used to produce large amount of heat at that point.

Question 9:_ One half of a convex lens is covered with a black paper. Will this lens produce a complete image of an object? Verify your answer experimentally. Explain your observations.

Answer:_ The convex lens will form complete image of an object, even if its one half is covered with black paper. It can be understood by the following two cases.

Case I: When the upper half of the lens is covered.

In this case a ray of light coming from the object will be reflected 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 as shown in following figure.

Case II: When the lower half of lens is convered.

In this case, a ray of light coming from the object is 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 as shown in the following figure.

Question 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 nature of the image formed.

Answer:_ Object distance u = ─ 25 cm

Object height = 5 cm

Focal length f = + 10 cm

According to lens formula

The position value of v shows that the image is formed at the other side of the lens.

The negative sign shows that the image is real and formed behind the lens.

H₁= m × H_o = ─ 0•66 × 5 = ─ 3•3 cm.

The negative value of image height indicate that image formed in inverted. The position, size and nature of image are shown in following ray diagram.

Question 11:_ A concave lens of focal length 15 cm from an image 10 cm from the lens. How far is the object place from the lens? Draw the diagram.

Answer:_ Focal length of concave lens ( OF₁) f = ─ 15 cm

Image distance, v = ─ 10 cm

According to lens formula

The negative value of the u indicates that the object is placed 30 cm in front of the lens. This is shown in the following ray diagram.

Question 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:_ Focal length of convex mirror, ( f ) = + 15 cm

Object distance (u) = ─ 10 cm

According to mirror formula

The +ve value of v indicates that the image formed behind the mirror.

m = + 0・6

The +ve value of magnification indicates that image formed is virtual and erect.

Question 13:_ The magnification produced by a plane mirror is + 1. What does this mean?

Answer:_ We know that

Since the magnification produced is +1. It means the image formed by the plane mirror is of the same size as that of the object. The positive side show that image formed is virtual and erect.

Question 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:_ Object distance (u) = ─ 20 cm

Object Height (h) = ─ 5 cm

Radius of curvature (R) = ─30 cm

Since, Radius of curvature = 2 × Focal Length

\ R = 2 × f

⟹ f = 30 ⁄ 2 = 15 cm i.e f = + 15 cm

According to mirror formula

\ v = 8・57 cm

The positive value of v indicates that the mirror is formed behind the mirror.

The positive value of magnification indicate that the image formed is virtual.

h^/ = m × h = 0•428 × 5 = 2•14 cm ⟹ h^/ = 2•14 cm
The positive value of magnification indicate that the image formed is erect. Therefore the image formed is virtual, erect and smaller in size.

Question 15:_ An object of size 7 cm is placed at 27 cm in front of the concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed. So that sharp focused image can be obtained? Find the size and nature of the image.

Answer:_ Object distance (u) = ─ 27 cm

Object Height (h) = 7 cm

Focal Length f = ─ 18 cm

According to mirror formula

⟹ v = ─ 54 cm

The screen should be placed at a distance of 54 cm in front of the given mirror.

The negative value of magnification indicates that the image formed is real.

h^/ = m × h = 7 × ─ 2 = ─ 14 cm ⟹ h^/ = ─ 14 cm

The negative sign show that image formed is inverted.

Question 16:_ Find the focal length of power ─ 20 D. What types of lens is this?

Answer:_

A concave lens has a negative focal length. Hence it is a concave lens.

Question 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:_

Hence the lens is convex or a converging lens.

More Questions

Question 1:_ Define the principal focus of a concave mirror.

Answer: _ Light ray that are parallel to the principal axis of a concave mirror converge at a specific point on its principal axis after reflection from the mirror. This point is known as principal focus of the concave mirror.

Question 2:_ The radius of curvature of a spherical mirror is 20 cm. What is its focus length?

Answer:_ Radius of curvature R = 20 cm

As we know that

Radius of curvature = 2 × Focal length

Hence, the focal length of the given spherical mirror is 10 cm.

Question 3:_ Name the mirror that can give an erect and enlarged image of an object.

Answer:_ Concave mirror produce an erect and enlarged image when the object is placed between its pole and principal focus.

Question 4:_ Why do we prefer a convex mirror as a rear view mirror in vehicles?

Answer:_ Concave mirror are prefer as a rear view mirror in vehicles because of the following reasons:

(1) Convex mirror produce an erect image of the object (traffic behind).

(2) Convex mirror give a wider field of view.

(3) The size of the image is much smaller than the size of the object which allows the driver to see most of the traffic behind him.

Question 5:_ Find the focal length of the convex mirror whose radius of curvature is 32cm.

Solution:_

Given Radius of curvature = + 32 cm (for convex mirror R is +ve)

Hence the focal length of the given convex mirror is 16 cm.

Question 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: _ For real image m is taken as ─ ve

So, the image is formed at 30 cm on the same side of the mirror as that of an object.

Question 7_: A ray of light travelling in air enters obliquely into water. Does the light ray bend toward normal or away from normal? Why?

Answer. The light ray bends toward normal.

When a ray of light travels from an optically rarer medium to an optically denser medium, its speed slow down and it bends towards normal. Since water is optical denser than air. Therefore a ray of light travelling from air into water will bends toward the normal.

Question 8:_ Light enters from air to glass having refractive index 1•50. What is the speed of light in the glass? The speed of light in vacuum is 3 × 10⁸ m/s.

Answer: _ We know that the refractive index of the medium n_m is given

Here, speed of light = 3 × 10⁸ m/s (given)

Refractive index = 1•50

Speed of light in glass

\ Speed of light in glass is 2 × 10⁸ m/s

Question 9:_ Find out from table the medium having highest optical density. Also find the median with lowest optical density.

Material Medium	Refractive Index	Material Medium	Refractive Index
Air	1.003	Crown glass	1.52
Ice	1.31	Canada Balsam	1.53
Water	1.33	Rock Salt	1.54
Alcohol	1.36	Carbon Disulphate	1.63
Kerosene	1.44	Dense flint glass	1.65
Fused quartz	1.46	Ruby	1.71
Turpentine Oil	1.47	Sapphire	1.71
Benzene	1.50	Diamond	2.42

Answer:_ Optical density of a medium depends upon its refractive index. The higher the refractive index, higher the optical density and vice versa.

From the table it is appearant that diamond is having highest optical density (n = 2.42)) and air is having lowest optical density (n = 1.003).

Question 10:_ You are given kerosene, turpentine and water. In which of these does the light travels faster? Use the information given in the table.

Answer:_ Refractive index of kerosene, turpentine and water are 1.44, 1.47, 133 respectively.

As we know that the velocity of light in light in medium is inversely proportional to the refractive index of the medium.

Therefore, the light will travel faster in water having refractive index i.e. n = 1.33.

Question 11:_ The refractive index of diamond is a 2.42. What do you mean by this statement?

Answer:_ Refractive index show the comparison of light speed in two mediums.

Where C is the speed of light in air.

The refractive index of diamond is 2.42. this suggest that the speed of light in diamond will reduce by a factor 2.42 compared to the speed of light in air.

Question 12:_ Define 1 dioptre of Power.

Answer:_ Power of Lens is reciprocal of its focal length. if P is a power of lens of focal length F in metre then

The S.I unit of the power of lens is dioptre. It is denoted by D. 1 Dioptre is defined as a power of lens of focal length 1 metre.

∴ D = 1 m^{─ 1}

Question 13:_ A concave lens form a real and inverted image of a needle at a distance of 50 cm in front of it. Where is the needle placed in front of convex lens if the image is equal to the size of the object? Also find the power of the lens.

Answer:_ When an object is placed at the centre of curvature 2F₁ of a convex lens, its image is formed at the centre of the curvature 2F₂ on the other side of the lens. The image formed is inverted and of the same size as the object, as shown in figure.

It is given that image of needle is formed at the distance of 50 cm from the convex lens.

Hence the needle is place in the front of the lens at a distance of 50 cm .

Object distance u = ─ 50 cm

Image distance v = 50 cm

Focal length = f

According to the lens formula

⟹ f = 25 cm = 0.25 m

Hence the power of the given lens is + 4 D.

Question 14:_ Find the power of convex lens of the focal length 2 m.

Answer:_ Here given focal length f = ─2 m

Here negative sign show the divergent nature of concave lens. Hence the power of the given concave lens is ─ 0.5D i.e. P = ─ 0.5 D