Question 1:
Figure 6.28 shows the path of light which travels from air to a prism.
(a) Determine the angle of refraction of light ray in the prism. The refractive index of the prism is 1.50.
(b) Will the light ray experience total internal reflection at the side AB of the prism? Explain your answer.
Answer:
(a)
$$ \begin{aligned} &\begin{aligned} n & =\frac{\sin i}{\sin r} \\ \sin r & =\frac{\sin i}{n} \\ & =\frac{\sin 45^{\circ}}{1.50} \\ r & =\sin ^{-1}\left(\frac{\sin 45^{\circ}}{1.50}\right) \end{aligned}\\ &\text { Angle of refraction, } r=28.1^{\circ} \end{aligned} $$
(b)
Figure 6.28 shows the path of light which travels from air to a prism.
(a) Determine the angle of refraction of light ray in the prism. The refractive index of the prism is 1.50.
(b) Will the light ray experience total internal reflection at the side AB of the prism? Explain your answer.
Answer:
(a)
$$ \begin{aligned} &\begin{aligned} n & =\frac{\sin i}{\sin r} \\ \sin r & =\frac{\sin i}{n} \\ & =\frac{\sin 45^{\circ}}{1.50} \\ r & =\sin ^{-1}\left(\frac{\sin 45^{\circ}}{1.50}\right) \end{aligned}\\ &\text { Angle of refraction, } r=28.1^{\circ} \end{aligned} $$
(b)
The phenomenon of total internal reflection happens when light travels from a medium of high optical density to a medium of low optical density, with the angle of incidence, i larger than the critical angle, c.
$$ \begin{aligned} & n=\frac{1}{\sin c} \\ & \mathrm{c}=\sin ^{-1}\left(\frac{1}{n}\right) \\ & \mathrm{c}=\sin ^{-1}\left(\frac{1}{1.5}\right) \\ & \mathrm{c}=41.8^{\circ} \\ & i=16.9^{\circ}, \quad c=41.8^{\circ} \\ & 16.9^{\circ}<41.8^{\circ} \\ & i<\mathrm{c} \end{aligned} $$Since the angle of incidence is smaller than the critical angle, thus, total internal reflection does not happen.
Question 2:
Figure 6.29 shows a fine optical fibre made up of inner core which has a high refractive index surrounded by cladding material of a low refractive index.
(a) Determine the critical angle of the optical fibre.
(b) What is the advantage of fine optical fibre?
Answer:
(a)
$$ \begin{aligned} \sin c & =\frac{1}{n} \\ \sin c & =\frac{1}{1.8} \\ c & =\sin ^{-1}\left(\frac{1}{1.8}\right) \\ & =33.75^{\circ} \end{aligned} $$
(b)
Advantages of fine optical fibre:
– It is light and easy to carry.
– It is used in a fibre-optic system such as endoscope, a medical instrument that is used to view internal parts of the human body.
– It is also used to transmit information in telephone and data signals at high speed, thus enabling multiple telephone messages and data to be sent simultaneously.
Figure 6.29 shows a fine optical fibre made up of inner core which has a high refractive index surrounded by cladding material of a low refractive index.
(a) Determine the critical angle of the optical fibre.
(b) What is the advantage of fine optical fibre?
Answer:
(a)
$$ \begin{aligned} \sin c & =\frac{1}{n} \\ \sin c & =\frac{1}{1.8} \\ c & =\sin ^{-1}\left(\frac{1}{1.8}\right) \\ & =33.75^{\circ} \end{aligned} $$
(b)
Advantages of fine optical fibre:
– It is light and easy to carry.
– It is used in a fibre-optic system such as endoscope, a medical instrument that is used to view internal parts of the human body.
– It is also used to transmit information in telephone and data signals at high speed, thus enabling multiple telephone messages and data to be sent simultaneously.