
An optical fibre is made of quartz filaments of refractive index 1.70 and has a coating of a material whose refractive index is 1.45 the range of angle of incidence for one laser beam to suffer total internal reflection is:
A) ${0^ \circ }$ to ${56.8^ \circ }$
B) ${0^ \circ }$ to ${62.6^ \circ }$
C) ${0^ \circ }$ to ${90^ \circ }$
D) ${0^ \circ }$ to ${180^ \circ }$
Answer
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Hint: The phenomenon of Total internal reflection only occurs when the angle of incidence is greater than the critical angle. These types of questions can be solved by first finding the critical angle and then using Snell’s law to obtain the angle of incidence.
Formula Used:
Snell’s law is used to obtain the angle of incidence. Mathematically it is given as follows:
$ \mu = \dfrac{{\sin i}}{{\sin r}} $
Complete step by step solution:
Refractive index of quartz filaments is 1.70 and the refractive index of coating material is 1.45. Thus, total refractive index is given by
$ \mu = \dfrac{{1.70}}{{1.45}} = 1.17 $
Now we know that the sine of a critical angle is the inverse of the refractive index of the medium. Thus,
$ \sin {i_c} = \dfrac{1}{\mu } = \dfrac{1}{{1.17}} = 0.856 $
Now we can find the value of critical angle.
$ {i_c} = {\sin ^{ - 1}}0.856 = 58.5^\circ $
We know that for total internal reflection to take place, the value of angle of incidence should be more than that of the critical angle. Thus,
$ {i'} > 58.5^\circ $
So, angle of refraction is given as:
$ r = 90 - {r'} $
Here r is the angle of refraction of the laser beam and r’ is the angle of refraction inside the fibre. Thus,
$
r < {90^ \circ } - 58.5^\circ \\
\Rightarrow r < 31.5^\circ $
Now we can use Snell’s law to obtain an angle of incidence. According to Snell’s law:
$\dfrac{{\sin i}}{{\sin r}} = {}_g^a\mu $
Putting the respective values in the above equation, we get:
$\dfrac{{\sin i}}{{\sin 31.5^\circ }} = 1.70$
Now, we can easily solve the equation.
$
\sin i = 1.70 \times \sin 31.5^\circ \\
\Rightarrow \sin i = 1.70 \times 0.524 = 0.89$
Now we can easily find the value of angle of incidence.
$ i = {\sin ^{ - 1}}0.89 = 62.6^\circ $
Thus, the value of angle of incidence can be anywhere between 0° to 62.6°. Hence the correct option is option (B).
Note: Usually students get confused when there are multiple angles of reflections and angle of incidences. It is important to use the value of angle of incidence and reflection correctly. Always remember to match the values according to the value of the refractive index.
Formula Used:
Snell’s law is used to obtain the angle of incidence. Mathematically it is given as follows:
$ \mu = \dfrac{{\sin i}}{{\sin r}} $
Complete step by step solution:
Refractive index of quartz filaments is 1.70 and the refractive index of coating material is 1.45. Thus, total refractive index is given by
$ \mu = \dfrac{{1.70}}{{1.45}} = 1.17 $
Now we know that the sine of a critical angle is the inverse of the refractive index of the medium. Thus,
$ \sin {i_c} = \dfrac{1}{\mu } = \dfrac{1}{{1.17}} = 0.856 $
Now we can find the value of critical angle.
$ {i_c} = {\sin ^{ - 1}}0.856 = 58.5^\circ $
We know that for total internal reflection to take place, the value of angle of incidence should be more than that of the critical angle. Thus,
$ {i'} > 58.5^\circ $
So, angle of refraction is given as:
$ r = 90 - {r'} $
Here r is the angle of refraction of the laser beam and r’ is the angle of refraction inside the fibre. Thus,
$
r < {90^ \circ } - 58.5^\circ \\
\Rightarrow r < 31.5^\circ $
Now we can use Snell’s law to obtain an angle of incidence. According to Snell’s law:
$\dfrac{{\sin i}}{{\sin r}} = {}_g^a\mu $
Putting the respective values in the above equation, we get:
$\dfrac{{\sin i}}{{\sin 31.5^\circ }} = 1.70$
Now, we can easily solve the equation.
$
\sin i = 1.70 \times \sin 31.5^\circ \\
\Rightarrow \sin i = 1.70 \times 0.524 = 0.89$
Now we can easily find the value of angle of incidence.
$ i = {\sin ^{ - 1}}0.89 = 62.6^\circ $
Thus, the value of angle of incidence can be anywhere between 0° to 62.6°. Hence the correct option is option (B).
Note: Usually students get confused when there are multiple angles of reflections and angle of incidences. It is important to use the value of angle of incidence and reflection correctly. Always remember to match the values according to the value of the refractive index.
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