An aquarium filled with water has a flat glass surface. A beam of light strikes the glass surface from outside making angle 90 degrees to the side of the aquarium(shown in the fig).a)(i)What is the angle of refraction when the ray passes from glass to water? (ii)Calculate the refractive index of the aquarium glass surface if the speed of light becomes 68% of its previous value while passing through the glass material.3b) The Following figure depicts a ray of light traveling from air into several mediums. Rank the mediums according to the refractive index from greatest to least. Justify your answer.
Answers
Explanation:
A common Physics lab is to sight through the long side of an isosceles triangle at a pin or other object held behind the opposite face. When done so, an unusual observation - a discrepant event - is observed. The diagram on the left below depicts the physical situation. A ray of light entered the face of the triangular block at a right angle to the boundary. This ray of light passes across the boundary without refraction since it was incident along the normal (recall the If I Were An Archer Fish page). The ray of light then travels in a straight line through the glass until it reaches the second boundary. Now instead of transmitting across this boundary, all of the light seems to reflect off the boundary and transmit out the opposite face of the isosceles triangle. This discrepant event bothers many as they spend several minutes looking for the light to refract through the second boundary. Then finally, to their amazement, they looked through the third face of the block and clearly see the ray. What happened? Why did light not refract through the second face?
The phenomenon observed in this part of the lab is known as total internal reflection. Total internal reflection, or TIR as it is intimately called, is the reflection of the total amount of incident light at the boundary between two media. TIR is the topic of focus in Lesson 3.
To understand total internal reflection, we will begin with a thought experiment. Suppose that a laser beam is submerged in a tank of water (don't do this at home) and pointed upwards towards water-air boundary. Then suppose that the angle at which the beam is directed upwards is slowly altered, beginning with small angles of incidence and proceeding towards larger and larger angles of incidence. What would be observed in such an experiment? If we understand the principles of boundary behavior, we would expect that we would observe both reflection and refraction. And indeed, that is what is observed (mostly). But that's not the only observation that we could make. We would also observe that the intensity of the reflected and refracted rays do not remain constant. At angle of incidence close to 0 degrees, most of the light energy is transmitted across the boundary and very little of it is reflected. As the angle is increased to greater and greater angles, we would begin to observe less refraction and more reflection. That is, as the angle of incidence is increased, the brightness of the refracted ray decreases and the brightness of the reflected ray increases. Finally, we would observe that the angles of the reflection and refraction are not equal. Since the light waves would refract away from the normal (a case of the SFA principle of refraction), the angle of refraction would be greater than the angle of incidence. And if this were the case, the angle of refraction would also be greater than the angle of reflection (since the angles of reflection and incidence are the same). As the angle of incidence is increased, the angle of refraction would eventually reach a 90-degree angle. These principles are depicted in the diagram below.
Answer:
26.9 , 31.9 , 41.1
Explanation:
Given Data
The index of refraction of the glass is:
n
g
=
1.52
.
The incidence angle of the light beam on the glass is:
θ
=
43.5
∘
.
The incident angle for part (c) is:
α
=
90
∘
.
(a)
Apply the Snell's law between air and glass to calculate the refracted angle in the glass.
n
a
sin
θ
=
n
g
sin
β
Here,
n
a
is the index of refraction for air and its value is
1
.
Substitute all the values in the above expression.
(
1
)
(
sin
43.5
∘
)
=
1.52
(
sin
β
)
β
=
26.9
∘
Thus, the refracted angle in the glass is
26.9
∘
.
(b)
Apply the Snell's law between water and glass to calculate the refracted angle in the glass.
n
w
sin
ϕ
=
n
g
sin
β
Here,
n
w
is the index of refraction for water and its value is
1.3
.
Substitute all the values in the above expression.
(
1.3
)
(
sin
ϕ
)
=
1.52
(
sin
26.9
∘
)
ϕ
=
31.9
∘
Thus, the angle of refraction in water is
31.9
∘
.
(c)
Apply the Snell's law between air and glass to calculate the refracted angle in the glass.
n
a
sin
α
=
n
g
sin
γ
Substitute all the values in the above expression.
(
1
)
(
sin
90
∘
)
=
1.52
(
sin
γ
)
γ
=
41.1
∘
Apply the Snell's law between water and glass to calculate the refracted angle in the glass.
n
w
sin
δ
=
n
g
sin
γ
Substitute all the values in the above expression.
(
1.3
)
(
sin
δ
)
=
1.52
(
sin
41.1
∘
)
δ
=
50.2
∘