Complete Question
You can determine the index of refraction of a substance by determining its critical angle. (a) What is the index of refraction of a substance that has a critical angle of [tex]46^0[/tex] when submerged in water ([tex]n_{water} = 1.333[/tex])? (b)What would the critical angle be for this substance in air([tex]n_{air} = 1.000293[/tex])
Your answer should be a number with three decimal places, do not include the unit.
Answer:
a
[tex]n_1 = 1.8489[/tex]
b
[tex]\theta _c = 32.75^o[/tex]
Explanation:
From question we are told that
The critical angle is [tex]\theta = 46^o[/tex]
The refractive index of water is [tex]n_{water} = 1.333[/tex]
Generally the critical angle is mathematically represented as
[tex]\theta _c = sin^{-1}[ {\frac{n_2}{n_1} }][/tex]
Considering question a
Here [tex]n_2 = n_{water } = 1.333[/tex]
[tex]\theta_c = \theta = 46^o[/tex]
So
[tex]n_1 = \frac{ n_{water}}{sin(\theta_c)}[/tex]
=> [tex]n_1 = \frac{ 1.33 }{sin(46)}[/tex]
=> [tex]n_1 = 1.8489[/tex]
Considering question b
Generally the critical angle is mathematically represented as
[tex]\theta _c = sin^{-1}[ {\frac{n_2}{n_1} }][/tex]
Here [tex]n_2 = n_{air } = 1.000293[/tex]
[tex]\theta _c = sin^{-1}[ {\frac{1.000293}{1.8489} }][/tex]
=> [tex]\theta _c = 32.75^o[/tex]
