Answer:
To find the speed of light in the plastic material, we can use Snell's law, which relates the refractive indices of two materials and the angles of incidence and refraction.
Snell's Law is given by:
�
1
×
sin
(
�
1
)
=
�
2
×
sin
(
�
2
)
n
1
×sin(θ
1
)=n
2
×sin(θ
2
)
Where:
�
1
n
1
and
�
2
n
2
are the refractive indices of the first and second mediums respectively,
�
1
θ
1
is the angle of incidence,
�
2
θ
2
is the angle of refraction.
Given that the critical angle for light in the plastic material placed in air is 37º, this means that light travels along the interface between air and the plastic material at an angle of 90º - 37º = 53º with respect to the normal.
Let's denote:
�
air
n
air
as the refractive index of air (which is approximately 1),
�
plastic
n
plastic
as the refractive index of the plastic material.
We can rearrange Snell's Law to solve for the refractive index of the plastic material:
�
plastic
=
�
air
×
sin
(
�
2
)
sin
(
�
1
)
n
plastic
=
sin(θ
1
)
n
air
×sin(θ
2
)
Substituting the given values:
�
plastic
=
1
×
sin
(
3
7
∘
)
sin
(
5
3
∘
)
n
plastic
=
sin(53
∘
)
1×sin(37
∘
)
�
plastic
≈
sin
(
3
7
∘
)
sin
(
5
3
∘
)
n
plastic
≈
sin(53
∘
)
sin(37
∘
)
Now, since
�
plastic
=
�
air
�
plastic
n
plastic
=
c
plastic
c
air
, where
�
air
c
air
and
�
plastic
c
plastic
are the speeds of light in air and plastic respectively, we can write:
�
plastic
=
�
air
�
plastic
c
plastic
=
n
plastic
c
air
�
plastic
=
3.0
×
1
0
8
sin
(
3
7
∘
)
sin
(
5
3
∘
)
c
plastic
=
sin(53
∘
)
sin(37
∘
)
3.0×10
8
Let's calculate this:
�
plastic
=
3.0
×
1
0
8
sin
(
3
7
∘
)
sin
(
5
3
∘
)
≈
3.0
×
1
0
8
0.6018
/
0.7992
c
plastic
=
sin(53
∘
)
sin(37
∘
)
3.0×10
8
≈
0.6018/0.7992
3.0×10
8
�
plastic
≈
3.0
×
1
0
8
0.7523
c
plastic
≈
0.7523
3.0×10
8
�
plastic
≈
3.986
×
1
0
8
m/s
c
plastic
≈3.986×10
8
m/s
So, the speed of light in the plastic material is approximately
3.986
×
1
0
8
3.986×10
8
m/s.
The closest option is C)
3.8
×
1
0
8
3.8×10
8
m/s.
Explanation:
