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In searching the bottom of a pool at night, a watchman shines a narrow beam of light from his flashlight, 1.3 m above the water level, onto the surface of the water at a point d = 2.8 m from his foot at the edge of the pool a) Where does the spot of light hit the bottom of the h = 2.1-m-deep pool? Measure from the bottom of the wall beneath his foot.

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The horizontal distance will be "4.6 m".

According to the question,

  • Length, [tex]l_1 = 2.1 \ m[/tex]
  • Height, [tex]h_1 = 1.3 \ m[/tex]
  • Distance, [tex]d = 2.8 \ m[/tex]

The angle of incidence will be:

→ [tex]tan \Theta_1= \frac{l_1}{h_1}[/tex]

             [tex]= \frac{2.1}{1.3}[/tex]

             [tex]= 2.076[/tex]

then,

        [tex]\Theta_1 = 64.3^{\circ}[/tex]        

From air into water, we get

→       [tex]n_{air} Sin \Theta_1 = n_{water} Sin \Theta_2[/tex]

  [tex](1.00) Sin 64.3^{\circ} = (1.3) Sin \Theta_2[/tex]

                     [tex]\Theta_2 = 42.6^{\circ}[/tex]

hence,

The horizontal distance,

→ [tex]l_2 = l_1+h_2 tan \Theta[/tex]

By substituting the values, we get

      [tex]= 2.1+(2.1)tan 42.6[/tex]

      [tex]= 4.6 \ m[/tex]

Thus the above approach is right.

Learn more about distance here:

https://brainly.com/question/22371581

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