What is the Law of Reflection?
There's a lot to be understood about the law of reflection. We will talk about the angle of reflection, the angle of incidence, the incident ray itself, and the planes all of this occurs on, but before we get into it, let us talk about what exactly is reflection.
Reflection is the sole reason why we can see "objects". The bouncing back of the light after it falls on a smooth surface is known as reflection; to be concise, the process of the light bouncing back after it strikes a surface is known as reflection.
Also, keep in mind, the incident ray, the reflected ray, and the normal ray lie and exist on the same plane.
The law of reflection helps study the images produced by plane and curved mirrors.
The formula for Law of Reflection:
| θi = θr |
Here, the angle of incidence is depicted using θi
and, the angle of reflection is depicted using θr |
Image source - ncert
There are two main types of reflections:
- Regular Reflection – As the name suggests, a regular reflection includes plane mirrors and a smooth surface. When the light bounces off of a plane mirror with a smooth surface, the image created is clear and much more visible.
- Irregular Reflection – As the name suggests, this kind of reflection occurs at a more irregular level where the rays bounce off in irregular directions covering more areas. Unlike the reflection mentioned above, which occurs on smooth surfaces, this reflection occurs on natural surfaces, which are often more than not, irregular and rough.
Things to understand about the Laws of Reflection
The fundamental concept to understand here is that when a ray/beam of light falls on a smooth surface, at that given time, the angle of reflection, i.e., θr, and the angle of incidence, i.e., θi are equal. One more important point is that the reflected ray and incident ray are normal to the surface and notably lie in the same plane.
Laws of Reflection for Class 11
The chapter 'optics' holds a weightage of 7 marks. It consists of one short question of 2 marks and one long question of 5 marks.
Illustrated examples
Example 1) Write and illustrate the formula of laws of reflection.
Answer – The formula of laws of reflection is θi = θr. The illustration is as follows:
Image source - ncert
Example 2) Illustrate the converging action of a convex lens with respect to the laws of reflection.
Answer - The illustration of the converging action of a convex lens is shown below:
Image source - ncert
Example 3) Illustrate Regular reflection.
Answer – A regular reflection includes plane mirrors and a smooth surface. When the light bounces off of a plane mirror with a smooth surface, the image created is clear and much more visible.
Image source - ncert
FAQs on Laws of Reflection
Q: Figure 9.32 shows a cross-section of a ‘light pipe’ made of a glass fibre of refractive index 1.68. The outer covering of the pipe is made of a material of refractive index 1.44. What is the range of the angles of the incident rays with the axis of the pipe for which total reflections inside the pipe take place, as shown in the figure. (b) What is the answer if there is no outer covering of the pipe?
A: Refractive index of the glass fibre, ?1 = 1.68
Refractive index of outer covering of the pipe, ?2 = 1.44
Given:
Angle of incidence = i, Angle of refraction = r, Angle of incidence at the interface = i’
The refractive index of the inner core-outer core interface, ? is given as
? = ?2/?1 = 1/sin ??′
1 / sin ??′ = 1.44 / 1.68 or sin ??′ = 1.44 / 1.68
i’ = 59°
For the critical angle, total reflection take place only when i > i’ i.e. i > 59°
Maximum angle of reflection, ???? = 90° − ??′ = 31°
Let ????? be the maximum angle of incidence and ???? be the maximum angle of reflection.
? = sin ????? / sin r??? or sin ??max
= ?1 × sin ????
sin ????? = 1.68 × ????31°
????? = 59.91° ≈ 60°
The entire rays incident at angles lying in the range of 0 < i < 60 will suffer total internal
reflection.
- If the outer covering is absent, then:
Refractive index of the outer pipe, ?1 = Refractive index of air = 1
For the angle of incidence, i = 90°, we can write Snell’s law at ‘air – pipe’ interface as :
???? ?? / sinr = ?2= 1.68
sin ? = sin ?? / 1.68 = sin 90° / 1.68 = 1 / 1.68
r = 36.53°
i’ = 90° − 36.53° = 53.47°
Since i’ > r, all incident rays will suffer total internal reflection.
Q: At what angle should a ray of light be incident on the face of a prism of refracting angle 60° so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is 1.524.
A: The incident, refracted and emergent rays associated with a glass prism ABC is shown in the adjoined figure.
Angle of the prism, ∠? = 60°
Refractive index of the prism, ? = 1.524
Let ∠??1 be the incident angle, ∠?1 be the refracted angle, ∠?2 be the incidence angle on face
AC and ∠? be the emergent angle from the prism = 90 °
According to Snell’s law, for face AC, we can have:
sin ? / sin ?2= ?
sin ?2 = sin ? / ? = sin 90° / 1.524= 0.656
∠?2 = 41°
From the Δ ABC, ∠? = ∠?1 + ∠?2 . Hence ∠?1 = 60° − 41° = 19°
According to Snell’s law, for face AB, we have
? = ???? ??1 / sin ?1 or sin ?? 1 = ? × sin ? = 1.524 × sin 19° = 0.496
∠??1 = 29.75 °
Hence the angle of incidence is 29.75 °
1) If the angle of reflection is 60°, what would the incidence angle be?
Answer: Since θi = θr, the angle of incidence would also be 60°
2) What are the types of reflections?
Answer: There are only two types of reflections – Regular Reflection and Irregular Reflection.
3) In what kind of reflection is the image formed and seen?
Answer: Images can be seen when light bounces off of a smooth surface with a plane mirror – which is the concept of regular reflection.
4) What is the law of reflection?
Answer: The law of reflection states that the angle of reflection and angle of incidence is equal when the light bounces back from a smooth surface.
5) What is irregular reflection?
Answer: This reflection occurs at a more irregular level where the rays bounce off in irregular directions covering more areas.
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