Snell's law:

$\overline{){{\mathbf{\eta}}}_{{\mathbf{1}}}{\mathbf{s}}{\mathbf{i}}{\mathbf{n}}{{\mathbf{\theta}}}_{{\mathbf{1}}}{\mathbf{=}}{{\mathbf{\eta}}}_{{\mathbf{2}}}{\mathbf{s}}{\mathbf{i}}{\mathbf{n}}{{\mathbf{\theta}}}_{{\mathbf{2}}}}$

**Part (a)**

It's observable that tan θ_{a} = L/d_{0}

Therefore, L = d_{0} tan θ_{a}

The expression for L is d_{0} tan θ_{a}.

**Part (b)**

The numerical value of L is:

A fisherman spots a fish underneath the water. It appears that the fish is *d*_{0} = 0.78 m under the water surface at an angle of *θ*_{a} = 68° with respect to the normal to the surface of the water. The index of refraction of water is *n _{w} *= 1.3 and the index of refraction of air is

**(a)** The perpendicular distance from the apparent position of the fish to the normal of the water surface shown in the figure is L. Express L in terms of tan θ_{a} and d_{0}. **(b)** Solve for the numerical value of L, in meters. **(c)** Express the sine of the angle θ_{w}, in terms of θ_{a}, n_{w}, and n_{a}. **(d)** Solve for the numerical value of in degrees. **(e)** Now assume that the real position of the fish is directly below the apparent position, as shown in the figure. Express the real depth of the fish, d, in terms of L and tanθ_{w}.

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