Question

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

Answer 1

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

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

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

=>     $\theta_c = 53.05^o$