(a) Figure 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?

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Answer 1

(a)
$μ_{1} = 1.68, μ_{2} = 1.44$
$μ_{1} / μ_{2} = 1 / s i n i_{1} \Rightarrow i_{1} = 5 9^{o}$
For critical angle, total internal reflection takes place when $i > i_{1}$
maximum angle of reflection is $90 - i_{1} = 3 1^{o} = r_{m a x}$
Let $i_{m a x}$ be the maximum angle of incidence.
$μ_{1} = s i n i_{m a x} / s i n r_{m a x} \Rightarrow i_{m a x} = 6 0^{o}$
Thus all the ray between $0^{o} < i < 6 0^{o}$ will suffer TIR.

(b)
Now, $μ_{1} = 1$ (air)
$s i n i / s i n r = μ_{2} / μ_{1} \Rightarrow r = 36 . 5^{o}$
$i = 90 - 36.5 = 53 . 5^{o}$
Since $i > r$ , all incident ray will suffer TIR

Answer 2

(a)

μ_{1} = 1.68, μ_{2} = 1.44

μ_{1} / μ_{2} = 1 / s i n i_{1} \Rightarrow i_{1} = 5 9^{o}

For critical angle, total internal reflection takes place when

i > i_{1}

maximum angle of reflection is

90 - i_{1} = 3 1^{o} = r_{m a x}

Let

i_{m a x}

be the maximum angle of incidence.

μ_{1} = s i n i_{m a x} / s i n r_{m a x} \Rightarrow i_{m a x} = 6 0^{o}

Thus all the ray between

0^{o} < i < 6 0^{o}

will suffer TIR.

(b)

Now,

μ_{1} = 1

(air)

s i n i / s i n r = μ_{2} / μ_{1} \Rightarrow r = 36 . 5^{o}

i = 90 - 36.5 = 53 . 5^{o}

Since

i > r

, all incident ray will suffer TIR