Answer:
The angle between the emergent blue and red light is [tex]0.566^{o}[/tex]
Explanation:
We have according to Snell's law
[tex]n_{1}sin(\theta _{i})=n_{2}sin(\theta _{r})[/tex]
Since medium from which light enter's is air thus [tex]n_{1}=1[/tex]
Thus for blue incident light we have
[tex]1\times sin(40.05)=1.641\times sin(\theta _{rb})\\\\\therefore \theta _{rb}=sin^{-1}(\frac{sin(40.05)}{1.64})\\\\\theta _{rb}=23.10[/tex]
Similarly using the same procedure for red light we have
[tex]1\times sin(40.05)=1.603\times sin(\theta _{rr})\\\\\therefore \theta _{rr}=sin^{-1}(\frac{sin(40.05)}{1.603})\\\\\theta _{rr}=23.66^{o}[/tex]
Thus the absolute value of angle between the refracted blue and red light is
[tex]|23.66-23.10|=0.566^{o}[/tex]
The absolute value of angle between the refracted blue and red light is 0.58⁰.
The angle of the refracted light is determined by using Snell's law as shown below;
n₁sinθ₁ = n₂sinθ₂
For the blue incident light, the angle of refraction is calculated as follows;
[tex]sin\theta_2 = \frac{n_1sin\theta_1 }{n_2} \\\\\theta _2 = sin^{-1}(\frac{n_1sin\theta_1 }{n_2})\\\\\theta_2 = sin^{-1}(\frac{1 \times sin(40.05) }{1.641})\\\\\theta_2 = 23.09 \ ^0[/tex]
For the red incident light, the angle of refraction is calculated as follows;
[tex]\theta _2 = sin^{-1}(\frac{n_1sin\theta_1 }{n_2})\\\\\theta_2 = sin^{-1}(\frac{1 \times sin(40.05) }{1.603})\\\\\theta_2 = 23.67 \ ^0[/tex]
The absolute value of angle between the refracted blue and red light is calculated as;
|θ| = 23.67 - 23.09 = 0.58⁰.
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