Answer:
actual elevation angle above horizontal =
![90{\circ} - 64.232{\circ} = 25.768^(\circ)](https://img.qammunity.org/2020/formulas/physics/high-school/kb95y63gbnm0vllo9csg9alxd93mb7ke5m.png)
Given:
apparent angle =
![47.5^(\circ)](https://img.qammunity.org/2020/formulas/physics/high-school/u18ebc1euozi5zeej1bbcjzv47b0zdzqmz.png)
refractive index of water,
= 1.333
Solution:
angle of incidence,
![\angle i = 90{\circ} - 47.5{\circ} = 42.5^(\circ)](https://img.qammunity.org/2020/formulas/physics/high-school/2nkcr4g18iipoz5944x4q5b8x4nrszncp9.png)
refractive index of water,
= 1
Using Snell's law:
![\mu _(a)sini = \mu _(w)sinr](https://img.qammunity.org/2020/formulas/physics/high-school/yz9s05i8z35fpl6mt3ujbu2qk16sijqm2o.png)
![sinr = \mu _(w)sini = 1.333sin42.5^(\circ)](https://img.qammunity.org/2020/formulas/physics/high-school/futxfoiruqggs6xa28kudx7u280s5930lh.png)
![r =sin^(-1)0.90056 = 64.232^(\circ)](https://img.qammunity.org/2020/formulas/physics/high-school/k5ch7wk46ox20g31qluazbd1kb2kyjqchz.png)
angle of refraction,
![\angle r = 64.232^(\circ)](https://img.qammunity.org/2020/formulas/physics/high-school/j9tfuixjzs5ibclag337tr75dgv152n77b.png)
Now,
actual elevation angle above horizontal =
![90{\circ} - 64.232{\circ} = 25.768^(\circ)](https://img.qammunity.org/2020/formulas/physics/high-school/kb95y63gbnm0vllo9csg9alxd93mb7ke5m.png)