A plane wavefront $A_{1} B_{1}$ is incident at a boundary $A_{1} B_{2}$ as shown. It takes time $τ$ for the wavefront to travel from $B_{1}$ to $B_{2}$ . Speeds of propagation of light in medium $1$ and $2$ are $v_{1}$ and $v_{2}$ respectively and $v_{2} > v_{1}$ . For the total internal reflection of wavefront.

v_{1} τ > A_{1} B_{2}

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v_{2} τ > A_{1} B_{2}

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v_{1} τ < A_{1} B_{2}

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v_{2} τ < A_{1} B_{2}

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Solution

The correct option is A $v_{1} τ > A_{1} B_{2}$
From figure, in right-angled triangle $A_{1} B_{1} B_{2}$ ,
$(A_{1} B_{2})^{2} = (A_{1} B_{1})^{2} + (B_{1} B_{2})^{2}$ , by phythagoras theorem.
or, $(A_{1} B_{2})^{2} < (B_{1} B_{2})^{2} o r, A_{1} B_{2} < ν_{1} τ$

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The solution (x,y) for the given pair of linear equations is $a_{1}$ x + $b_{1}$ y + $c_{1}$ =0 and $a_{2}$ x + $b_{2}$ y + $c_{2}$ =0

For the given pair of linear equations $a_{1} x + b_{1} y + c_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} = 0,$ the solution $(x, y)$ is

(a_{1}, b_{1}), (a_{2}, b_{2})

[(a_{1} + a_{2}), (b_{1} + b_{2})]

a_{1} b_{2} = a_{2} b_{1}

Solve the two equations using method of elimination.

$a_{1} x + b_{1} y + c_{1} = 0$
$a_{2} x + b_{2} y + c_{2} = 0$

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