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SCORPION-xisa [38]

3 years ago

7

Find the angle of incidence at which a light beam traveling from a medium having the index of refraction n1 = 2.7 to another med

ium having the index of refraction n2 = 1.6 suffers a total internal reflection.

1 answer:

luda_lava [24]3 years ago

7 0

Answer:

The angle of incidence at which total internal reflection takes place will be equal to $\Theta _c=36.34^{\circ}$

Explanation:

We have given that light is traveling from a medium of refractive index x 2.7 to a medium of refractive index 1.6

So refractive index of first medium $n_1=2.7$ and $n_2=1.6$

We have to find the angle of incidence for total internal reflection

For total internal reflection

$n_1sin\Theta _c=n_2sin90^{\circ}$

$\Theta _c=sin^{-1}\frac{n_2}{n_1}$

$\Theta _c=sin^{-1}\frac{1.6}{2.7}$

$\Theta _c=36.34^{\circ}$

So the angle of incidence at which total internal reflection takes place will be equal to $\Theta _c=36.34^{\circ}$

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