A 4.00 m long pole stands vertically in a lake having a depth of 2.50 m. The Sun is 40.0° above the horizontal. Determine the length of the pole's shadow on the bottom of the lake. Take the index of refraction for water to be 1.33.

Answer:3.541 mStep-by-step explanation:  From Diagram we can say that[tex]tan\left ( 50\right )=\frac{b}{1.5}[/tex]b=1.787 malso we know[tex]\frac{sin50}{sin\theta}=\mu [/tex][tex]sin\theta =0.5745[/tex][tex]\theta =35.064^{\circ}[/tex][tex]also\ tan\theta =\frac{a}{2.5}[/tex]a=1.754thus shadow of pole[tex]=a+b=1.754+1.787=3.541 m[/tex]