
What should be the maximum acceptance angle at the air-core interface of an optical fibre if $ {n_1} $ and $ {n_2} $ are the refractive indices of the core and the cladding, respectively
A. $ {\sin ^{ - 1}}\left( {\dfrac{{{n_2}}}{{{n_1}}}} \right) $
B. $ {\sin ^{ - 1}}\sqrt {{n_1}^2 - {n_2}^2} $
C. $ {\tan ^{ - 1}}\left( {\dfrac{{{n_2}}}{{{n_1}}}} \right) $
D. $ {\tan ^{ - 1}}\left( {\dfrac{{{n_1}}}{{{n_2}}}} \right) $
Answer
524.7k+ views
Hint: Use the Snell’s law. Input the values of $ i $ and $ r $ at the air-core surface. Then obtain several equations, and use them to find the value of the acceptance angle. Alternatively, you can directly use the formula for acceptance angle if the derivation is not required.
Complete step by step solution:
Acceptance angle is defined as the maximum angle of incidence for which all incident light is totally reflected at the surface of a pair of media having some values of refractive indices.
The formula for acceptance angle is given by:
$ {\theta _{acc}} = {\sin ^{ - 1}}\left( {\dfrac{1}{{{n_0}}}\sqrt {{n_1}^2 - {n_2}^2} } \right) $
Here,
$ {\theta _{acc}} $ represents the acceptance angle.
$ {n_1} $ represents the refractive index of the core.
$ {n_2} $ represents the refractive index of the cladding.
$ {n_0} $ represents the refractive index of the medium surrounding the core and the cladding.
According to the question, the medium surrounding the core and the cladding is air. We all know that the refractive index of air is taken as one.
Putting $ {n_0} = 1 $ in the formula above, we get:
$
{\theta _{acc}} = {\sin ^{ - 1}}\left( {\dfrac{1}{1}\sqrt {{n_1}^2 - {n_2}^2} } \right) \\
{\theta _{acc}} = {\sin ^{ - 1}}\sqrt {{n_1}^2 - {n_2}^2} \\
$
We get the acceptance angle to be the inverse sine of the square root of the difference of the squares of the refractive indices of the core and the cladding.
Hence, Option (B) is the correct answer.
Note:
Snell’s law, in optics, is a relationship between the path taken by a ray of light in crossing the boundary or surface of separation between two contacting substances and the refractive index of each.
Complete step by step solution:
Acceptance angle is defined as the maximum angle of incidence for which all incident light is totally reflected at the surface of a pair of media having some values of refractive indices.
The formula for acceptance angle is given by:
$ {\theta _{acc}} = {\sin ^{ - 1}}\left( {\dfrac{1}{{{n_0}}}\sqrt {{n_1}^2 - {n_2}^2} } \right) $
Here,
$ {\theta _{acc}} $ represents the acceptance angle.
$ {n_1} $ represents the refractive index of the core.
$ {n_2} $ represents the refractive index of the cladding.
$ {n_0} $ represents the refractive index of the medium surrounding the core and the cladding.
According to the question, the medium surrounding the core and the cladding is air. We all know that the refractive index of air is taken as one.
Putting $ {n_0} = 1 $ in the formula above, we get:
$
{\theta _{acc}} = {\sin ^{ - 1}}\left( {\dfrac{1}{1}\sqrt {{n_1}^2 - {n_2}^2} } \right) \\
{\theta _{acc}} = {\sin ^{ - 1}}\sqrt {{n_1}^2 - {n_2}^2} \\
$
We get the acceptance angle to be the inverse sine of the square root of the difference of the squares of the refractive indices of the core and the cladding.
Hence, Option (B) is the correct answer.
Note:
Snell’s law, in optics, is a relationship between the path taken by a ray of light in crossing the boundary or surface of separation between two contacting substances and the refractive index of each.
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