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jeka94
3 years ago
6

A ray of light incident in water strikes the surface separating water from air making an angle of 10 ° with the normal to the su

rface. (refractive index of water = 1.3) a) What is the angle of refraction? b) What should be the angle of incidence if we want an angle of refraction not greater than 45 ° ? c) What is the critical angle?
Physics
1 answer:
labwork [276]3 years ago
4 0

Answer:

a

 \theta _2  = 13^o

b

 \theta _1  =32.94^o

c

 \theta_c  =  53.05^o    

Explanation:

From the question we are told that

    The angle of incidence is  \theta_1 =  10^o

    The refractive index of water is  n_1 = 1.3

  Generally Snell's law is mathematically represented as

          n_1 sin(\theta_1) =  n_2 sin(\theta_ 2)

Here n_2 is the refractive index of air with value  n_2 =  1

         \theta_2  is the angle of refraction

So  

        \theta _2  =  sin^{-1}[\frac{n_1 * sin(\theta _1)}{n_2} ]

=>     \theta _2  =  sin^{-1}[\frac{1.3 * sin(10)}{1} ]

=>     \theta _2  = 13^o

Given that the angle should not be greater than \theta _2 =45^o  then the angle of incidence will be

       \theta _1  =  sin^{-1}[\frac{n_2 * sin(\theta _2)}{n_1} ]

=>     \theta _1  =  sin^{-1}[\frac{1 * sin(45)}{1.3} ]

=>     \theta _1  =32.94^o

Generally for critical angle is mathematically represented as

        \theta_c  =  sin^{-1}[\frac{n_2}{n_1} ]

=>     \theta_c  =  sin^{-1}[\frac{1}{1.3} ]  

=>     \theta_c  =  53.05^o            

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