Answer:
Required horizontal distance is 0.9279 meters
Explanation:
The situation is represented in the attached figure
The horizontal distance can be seen to be equal to
[tex]H=H_{1}+H_{2}[/tex]
In the upper triangle we have
[tex]tan(45^{o})=\frac{0.52m}{H_{1}}\\\\\therefore 1=\frac{0.52}{H_{1}}\\\\\therefore H_{1}=0.52m[/tex]
Now the angle [tex]\lambda[/tex] can be calculated using Snell's law
By snell's Law we have
[tex]\mu _{1}sin(\theta _{1})=\mu _{2}sin(\theta _{2})[/tex]
Since light comes from air thus [tex]\mu _{1}=1[/tex]
Light enter's the water thus we have [tex]\mu _{2}=1.33[/tex]
Applying values we calculate [tex]\lambda[/tex] as
[tex]1\times sin(45^{o})=1.33sin(\lambda _{r})\\\\\lambda _{r}=sin^{-1}(\frac{sin(45^{o})}{1.33})\\\\\lambda _{r}=32.11^{o}[/tex]
Now in the attached figure we have
[tex]tan(90-\lambda _{r})=\frac{0.65}{H_{2}}\\\\\therefore tan(90-32.11)=\frac{0.65}{H_{2}}[/tex]
Solving for [tex]H_{2}[/tex] we get
[tex]H_{2}=\frac{0.65}{tan(90-32.11)}\\\\H_{2}=0.408m[/tex]
thus the required horizontal distance is [tex]0.52+0.408=0.927m[/tex]
