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Question: (a) Calculate the critical angle for glass air interface if a ray of light incident on a glass surfa...

(a) Calculate the critical angle for glass air interface if a ray of light incident on a glass surface is deviated through 15∘{{15}^{\circ }} when angle of incident is 45∘{{45}^{\circ }}
(b) A ray of light travels from denser medium having refractive index 2\sqrt{2} to air, what should be the angle of refracted when angle of incident is 30∘{{30}^{\circ }}

Explanation

Solution

Critical angle is the maximum angle of incidence by which the light travelling from denser to rarer medium strikes back. This concept is very useful in optical fiber where data or information is transmitted through light rays. The light travelling through optical fiber travels using Total Internal Reflection.

Complete answer:
A.) In this part we are given the angle of incidence as 45∘{{45}^{\circ }}, the question states that the ray deviated by 15∘{{15}^{\circ }} this means that the angle of reflection is reduced by 15∘{{15}^{\circ }}
We need to find the critical angle in this question, the formula of critical angle can be given as
sin⁡ic=1μ\sin {{i}_{c}}=\dfrac{1}{\mu }
Where μ\mu is the refractive index of the glass.
So, we can say that μ=sin⁡isin⁡r\mu =\dfrac{\sin i}{\sin r}
Therefore, the formula for critical angle now becomes
sin⁡ic=sin⁡rsin⁡i\sin {{i}_{c}}=\dfrac{\sin r}{\sin i} …………(1)
Now the angle of refraction can be given as ∠r=45∘−15∘\angle r={{45}^{\circ }}-{{15}^{\circ }}
Therefore, ∠r=30∘\angle r={{30}^{\circ }}
Now putting the value in equation (1), we get
sin⁡ic=sin⁡30sin⁡45\sin {{i}_{c}}=\dfrac{\sin 30}{\sin 45}
sin⁡ic=0.71\sin {{i}_{c}}=0.71
ic=sin⁡−1(0.71){{i}_{c}}={{\sin }^{-1}}(0.71)
ic=45.2∘{{i}_{c}}={{45.2}^{\circ }}
Therefore, the value of critical angle for the above situation is 45.2∘{{45.2}^{\circ }}
B.)
In the second part, we have been given that the angle if incidence is 30∘{{30}^{\circ }}
And the refractive index is 2\sqrt{2}
We know that the formula for refractive index is μ=sin⁡isin⁡r\mu =\dfrac{\sin i}{\sin r}
Putting the values in above formula we can find out the angle of refraction
2=sin⁡30∘sin⁡r\sqrt{2}=\dfrac{\sin {{30}^{\circ }}}{\sin r}
sin⁡r=0.353\sin r=0.353
r=sin⁡−1(0.353)r={{\sin }^{-1}}(0.353)
r=20.67∘r={{20.67}^{\circ }}
Hence, we can say that the angle of refraction is 20.67∘{{20.67}^{\circ }}

Note:
Whenever in a question we need to find out the angle of incidence or angle of refraction then always remember to use this formula μ=sin⁡isin⁡r\mu =\dfrac{\sin i}{\sin r}. This is very basic formula and it is used in most of the question in optics and you should try to use this formula to find out the remaining quantities in a given question