Question

When a light ray travels from a more dense medium to a less dense medium and the light ray intercepts the interface at some angle, the light ray will be deviated away from the normal to the interface.  When the angle of the incident light ray is equal or greater than some critical angle , the light ray will be bent away from the normal unit the angle of refraction becomes 90, and total internal reflection occurs.  The critical angle,  , for two media whose indices of refraction are n1 and n2 can be found from Snell's Law.

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A.)  If the index of refraction for the more dense medium is n1, so that n1>n2, show that the equation for the critical angle is given by      \(sin\theta{c}=n{2}/n{1}\)  .

B.)  Find the critical angle for water whose index of refraction is 1.33.

Answer 1

snells law says

$\frac{sin i}{sinr}=n2/n1$

r = 90 degrees

$\frac{sin i}{1}=n2/n1$

$sin\Theta c=n2/n1$

b) n2=1

n1=1.33

that gives

critical ange = 48.6 degrees