Optics in Swimming Pool

Solve it independently so that you understand this question More precisely !

---

Concept Used in This Question Is :

$\bullet$ Formulas for Paraxial Ray's does not hold good in this question

$\bullet$ For To Locate the image we require 2 Near by rays .

$\bullet$ Position Of Image Doesn't Change if we consider to near by rays .

---

Now Let horizontal distance of image from Pole "P" is y

So $\quad y\quad -\quad x\quad =\quad constant$ .

Now Let Two rays of angle of incidence $i\quad \quad \quad \& \quad \quad i\quad +\quad di$ .

Now stablished relation between them ( by using snell's Law )

and use $\cfrac { d(y\quad -\quad x) }{ di } \quad =\quad 0$ .

---

so we get

$h\quad =\quad \cfrac { 1{ (\cos { \theta } ) }^{ 3 } }{ \mu { H }^{ 2 } } { ({ H }^{ 2 }\quad +\quad { x }^{ 2 }) }^{ \cfrac { 3 }{ 2 } }$ .

---

NOTE:
Apparent depth will be maximum when we see Normally in swimming Pool.

Since $\theta =0$

direct from the well - known book

Thanks to you, couldn't solve there, but now I did.