Question

59. In a calm swimming pool, a person is viewing outside objects
by keeping an eye at a depth h inside water. If the critical
angle for water is , then the value of the diameter of the
circle of view for outside objects will be:
(a) 2h sin ,
(b) 2h cos e
(c) 2h tanec
(d) 2h cote
​

Answer 1

Given:

In a calm swimming pool, a person is viewing outside objects by keeping an eye at a depth h inside water. The critical angle for water is $\phi$

To find:

Diameter of the circle.

Calculation:

Applying SNELL'S LAW :

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

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

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

Cross multiplying and squaring on both sides :

$= > {r}^{2} { \mu}^{2} = {r}^{2} + {h}^{2}$

$= > {r}^{2} ( { \mu}^{2} - 1) = {h}^{2}$

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

Putting value of $\mu$ :

$= > r = \dfrac{h}{ \sqrt{ \frac{1}{ { \sin}^{2}( \phi) } - 1} }$

$= > r = \dfrac{h}{ \sqrt{ { cosec }^{2}( \phi) - 1} }$

$= > r = \dfrac{h}{ \sqrt{ { \cot }^{2}( \phi) } }$

$= > r = \dfrac{h}{ \cot( \phi) }$

$= > r = h \tan( \phi)$

So , diameter will be :

$\boxed{ \large{ \sf{ d = 2r = 2 h \tan( \phi) }}}$