Chapter 16 - Reflection of Light
1. Fill in the blanks
(i) The perpendicular to the mirror at the point of incidence is called…………
Ans: normal
(ii) The reflection of light from a wooden surface is………….. reflection.
Ans: irregular
(iii) The working of Kaleidoscope is based on the properties of …………..
Ans: reflection of light
2. Draw a figure describing the following.
The reflecting surfaces of two mirrors make an angle of 90⁰ with each other. If a ray incident of one mirror has an angle of incidence of 30⁰, draw the ray reflected from the second mirror. What will be its angle of reflection?
Ans:
For first mirror (PO):
Ray AB: Incident ray
Ray BC: Reflected ray
∠ABN₁: Angle of incidence
∠CBN₁: Angle of reflection
For second mirror (OQ):
Ray BC: Incident ray,
Ray CD: Reflected ray
∠BCN2: Angle of incidence
∠DCN2 : Angle of reflection
According to law of reflection,
∠ABN₁ = ∠CBN₁ = 30° … (i)
Since ∠CBN₁ and ∠BCO are alternate angles,
∴ ∠CBN₁ = ∠BCO = 30⁰….[From equation (i)] …(ii)
From figure, N₂ is normal to the second mirror,
∴ CN is perpendicular to OQ, ∠OCN₂ = 90⁰
∴ ∠BCO + ∠BCN₂ = 90⁰
∴ ∠BCN₂ = 90°- ∠BCO
∴ ∠BCN₂ = 90⁰ – 30⁰ ….[From equation (ii)]
∴ ∠BCN₂ = 60⁰
According to law of reflection,
∠N₂CD = ∠BCN₂ = 60⁰
Ans: The angle of reflection for the second mirror will be 60⁰
3. How will you explain the statement ‘we cannot see the objects in a dark room’?
Ans:
(i) The objects can be seen when light coming from the objects enters our eyes.
(ii) These light rays entering our eyes may be emitted by the object or reflected by the object.
(iii) As there is no light emitted or reflected in a completely dark room, we cannot see the objects.
4. Explain the difference between regular and irregular reflection of light.
Ans:
| Regular reflection of light | Irregular reflection of light |
|---|---|
(i) Reflection takes place from a plane and from a rough surface. | (i) Reflection takes place from a smooth surface. |
(ii) Angles of incidence for all parallel rays are the same. | (ii) Angles of incidence for all parallel rays are not the same. |
(iii) Reflected rays are parallel to one another. | (iii) Reflected rays are not parallel to one another. |
(iv) Angle of incidence and angle of reflection are equal at different points, i.e., i₁ = r₁ = i₂ = r₂…. | (iv) Angle of incidence and angle of reflection are not equal at different points, i.e., i₁ = r₁ ≠ i₂ = r₂…. |
5. Draw a figure showing the following.
a. Incident Ray
b. Normal
c. Angle of incidence
d. Angle of reflection
e. Point of incidence
f. Reflected ray
Ans:
Ray AO: Incident ray
Ray ON: Normal
∠AON: Angle of incidence
∠BON: Angle of reflection
Point O: Point of incidence
Ray OB: Reflected ray
6. Study the following incident.
Swara and Yash were looking in a water filled vessel. They could see their images clearly in the still water. At that instant, Yash threw a stone in the water. Now their images were blurred. Swara could not understand the reason for the blurring of the images.
Explain the reason for blurring of the images to Swara by answering the following question.
(i) Is there a relation between the reflection of light and the blurring of the images?
Ans: Yes, there is a relation between the reflection of light and blurring of the images.
(ii) Which types of reflection of light can you notice from this?
Ans: When water was still, regular reflection of light was noticed as the surface of water was smooth. As soon as Yash threw a stone in the water, the surface of the water became irregular. Hence, irregular reflection of light was noticed.
(iii) Are laws of reflection followed in these types of reflection?
Ans: Yes, laws of reflection are followed by both; regular and irregular types of reflection.
7. Solve the following examples.
a. If the angle between the plane mirror and the incident ray is 40⁰, what are the angles of incidence and reflection?
Given:
∠AOP = 40⁰ ….(i)
To find:
∠i and ∠r
Solution:
From figure,
∠AOP + ∠PON = 90°
∠PON = 90⁰ – ∠AOP
∠PON = 90°- 40°. ….[From (i)]
∴ ∠PON = ∠ i = 50°
According to law of reflection,
∠i = ∠r
∴ ∠r = 50⁰
Ans: The angle of incidence is 50⁰ and the angle of reflection is 50°.
b. If the angle between the mirror and reflected ray is 23⁰, what is the angle of incidence of the incident ray?
Given:
∠BOR = 23⁰ ….(i)
To find:
∠i
Solution:
From figure,
∠BOR + ∠RON = 90⁰
∠RON = 90⁰ – ∠BOR
∠RON = 90⁰ – 23⁰ ….[From (i)]
∴ ∠RON = angle r = 67⁰
According to law of reflection,
∠i = ∠r
∴ ∠i = 67⁰
Ans: The angle of incidence is 67⁰