A large square container with thin transparent vertical walls and filled with water (refractive index $\frac{4}{3}$ ) is kept on a horizontal table. A student holds a thin straight wire vertically inside the water 12 cm from one of its corners, as shown schematically in the figure. Looking at the wire from this corner, another student sees two images of the wire, located symmetrically on each side of the line of sight as shown. The separation (in cm) between these images is _______.

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Answer 1

Assume that observer see the image of object through edge $\Rightarrow a = 4 5^{\circ}$

$A B = \frac{1 2 d α}{cos α} = \frac{x d θ}{cos θ}$

Applying snell's law

$\frac{4}{3} sin a = 1 . s i n a$

$\frac{4}{3} cos a d a = cos θ d θ$

$\Rightarrow \frac{9}{cos ^{2} α} = \frac{x}{cos ^{2} θ}$

$1 . sin θ = \frac{4}{3} sin α$

$\Rightarrow sin θ = \frac{2 2}{3}$

$\Rightarrow x = 18 \times \frac{1}{9} = 2$

$d = 2 \times sin (θ - α)$

$= 4 \times \frac{1}{2} (\frac{2 2}{3} - \frac{1}{3}) = \frac{8 - 2 2}{3}$

$= 1.73 \approx 2$

Answer 2

Assume that observer see the image of object through edge

\Rightarrow a = 4 5^{\circ}

A B = \frac{1 2 d α}{cos α} = \frac{x d θ}{cos θ}

Applying snell's law

\frac{4}{3} sin a = 1 . s i n a

\frac{4}{3} cos a d a = cos θ d θ

\Rightarrow \frac{9}{cos ^{2} α} = \frac{x}{cos ^{2} θ}

1 . sin θ = \frac{4}{3} sin α

\Rightarrow sin θ = \frac{2 2}{3}

\Rightarrow x = 18 \times \frac{1}{9} = 2

d = 2 \times sin (θ - α)

= 4 \times \frac{1}{2} (\frac{2 2}{3} - \frac{1}{3}) = \frac{8 - 2 2}{3}

= 1.73 \approx 2