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# Let ray of light given by the vector $\overrightarrow{A}= 6 \sqrt 3 \hat i+8 \sqrt 3 \hat j -10 \hat k$ be incident on the $X-Y$ plane . Which is the boundary between two transparent medium. Medium 1 in $Z > 0$ has refractive index of $\sqrt 2$ and medium 2 with $Z > 0$ has refractive index of $\sqrt 3$. What is the angle of refraction in medium 2.

$(a)\;30^{\circ} \\ (b)\;60^{\circ} \\ (c)\;45^{\circ} \\ (d)\;15^{\circ}$

If the vector is represented by $a \hat i + b\hat j +c \hat k$
The angle it makes with z axis is given by
$\cos \theta _1 =\large\frac{|c|}{\sqrt {a^2 +b^2 +c^2}}$
over vector $= 6 \sqrt 3 \hat i + 8 \sqrt 3 \hat j -10 \hat k$
$\cos \theta_1=\large\frac{10}{\sqrt {(6 \sqrt 3 )^2 +(8 \sqrt 3)^2 +10^2}}$
$\qquad= \large\frac{10}{\sqrt {400}}$
$\qquad= \large\frac{10}{20}$
$\qquad= \large\frac{1}{2}$
$\therefore \theta_1= 60^{\circ}$
Now light travels from medium I of $\mu_1 = \sqrt {2}$ to medium II of $\mu_2= \sqrt 3$
$\therefore \mu_1 \sin \theta_1 =\mu_2 \sin \theta_2$
$\sqrt 2 \sin 60=\sqrt 3 \sin \theta _2$
$\sqrt 2 \times \large\frac{\sqrt 3 }{2}$$=\sqrt 3 \sin \theta _2$
$\sin \theta_2 =\large\frac{1}{\sqrt 2 }$
$\theta_2 =45^{\circ}$
Hence c is the correct answer.