Question

A disc is placed on the surface of a water lake of
refractive index ul. A fish is at a depth 'd below the
surface of water. What is the minimum area of the
disc so that the fish is not able to see the world
outside the lake?
​

Answer 1

Answer:

Given:

A fish is at depth of d from water surface. Refractive Index of water is μ.

To find:

The area of disc such that fish is not an to see anything outside water.

Diagram :

Please see the attached photo to understand better.

Calculation:

Applying Snell's Law :

$\therefore \mu \: \times \sin( \theta) = 1 \times \sin(90 \degree)$

$= > \sin( \theta) = \dfrac{1}{ \mu}$

$= > \dfrac{r}{ \sqrt{ {r}^{2} + {d}^{2} } } = \dfrac{1}{ \mu}$

$= > \: {( \mu r) }^{2} = {r}^{2} + {d}^{2}$

$= > {r}^{2} \{ { \mu}^{2} - 1 \} = {d}^{2}$

$= > {r}^{2} = \dfrac{ {d}^{2} }{ { \mu}^{2} - 1 }$

Now area of circular disc having radius r will be :

$\therefore \: area = \pi {r}^{2}$

$= > area = \dfrac{\pi {d}^{2} }{ { \mu}^{2} - 1 }$

So final answer :

$\: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \large{ \sf{ \blue{ \bold{ area = \dfrac{\pi {d}^{2} }{ { \mu}^{2} - 1 }}}}}}$

Answer 2

$\boxed{ \bold{ \sf{ \orange{ \huge{Answer}}}}}\\ \\ \star \sf \: \bold{ \red{GIVEN \: :}} \\ \\ \mapsto \sf \: Depth \: of \: fish \: from \: water \: surface = d \\ \\ \mapsto \sf \: Refractive \: index \: of \: water = \mu{ \tiny{w}} \\ \\ \mapsto \sf \: Refractive \: index \: of \: air = \mu{ \tiny{a}} \\ \\ \star \sf \: \bold{ \red{TO \: FIND \: :}} \\ \\ \mapsto \sf \: The \: minimum \: area \: of \: the \: disc \: so \\ \sf \: that \: the \: fish \: is \: not \: able \: to \: see \: the \\ \sf \: world \: outside \: the \: water. \\ \\ \star \sf \: \bold{ \red{CONCEPT \: :}} \\ \\ \mapsto \sf \: The \: fish \: is \: not \: able \: to \: see \: the \: outside \\ \sf \: in \: the \: case \: of \: critical \: angle \: view. \\ \\ \mapsto \sf \: In \: the \: condition \: of \: critical \: angle \\ \\ \sf \: angle \: of \: incident = \angle \: i = \theta{ \tiny{c}} \\ \sf \: angle \: of \: reflection = \angle \: r = 90 \degree \\ \\ \star \sf \: \bold{ \red{FORMULA \: DERIVATION \: :}} \\ \\ \mapsto \sf \: Appling \: snells \: law \: at \: water \: surface \\ \\ \mapsto \sf \: \mu{ \tiny{w}} \times sin\theta{ \tiny{c}} = \mu{ \tiny{a}} \times sin90 \degree \\ \\ \mapsto \sf \: sin \theta{ \tiny{c}} = \frac{1}{ \mu{ \tiny{w}}} = \frac{r}{ \sqrt{ {r}^{2} + {d}^{2} } } \\ \\ \sf \therefore \: r = \frac{d}{ \sqrt{ { (\mu{ \tiny{w}}})^{2} - 1} } \\ \\ \mapsto \sf \: Area \: of \: disc = \pi{ {r}^{2} } \\ \\ \therefore \: \underline{\boxed{ \bold{ \blue{ \sf{Area = \frac{ \pi{ {d}^{2} }}{ { (\mu{ \tiny{w}}})^{2} - 1 }}}}}}$