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

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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.

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