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After a long day of driving you take a late-night swim in a motel swimming pool. When you go to your room, you realize that you have lost your room key in the pool. You borrow a powerful flashlight and walk around the pool, shining the light into it. The light shines on the key, which is lying on the bottom of the pool, when the flashlight is held 1.2 m above the water surface and is directed at the surface a horizontal distance of 1.5 m from the edge (Fig. P33.44). If the water here is 4.0 m deep, how far is the key from the edge of the pool

Respuesta :

Answer:

Total length of pool [tex]4.4[/tex] meter

Explanation:

The image for the question is attached for better understanding

[tex]Tan \Theta= \frac{1.5}{1.2}\\\Theta_a = tan^{-1} (1.25)\\\Theta_a = 51.34[/tex]degree

Using the Snell’s law –

[tex]n_a * sin\Theta_a = n_b * sin\Theta_b\\1 * sin 51.34 = 1.33 * sin\Theta_b\\sin\Theta_b = 0587\\\Theta_b = 35.94\\Tan \Theta_b = \frac{BC}{4} \\BC = 4 * tan (35.94) \\BC = 2.9[/tex]

Total length of pool [tex]= 1.5 +2.9 = 4.4[/tex] meter

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Answer:

Explanation:

i is angle of incidence and r is angle of refraction .

Tan i = 1.5 / 1.2 = 1.25

i = 51.34⁰

refractive index of water = 1.33

sin i / sinr = 1.33

sin 51.34 / sin r = 1.33

sinr = .78 / 1.33 = .5864

r = 36⁰

From the figure given ,

AB / AP = Tan 36

AB = AP tan 36 = 4 x tan 36 = 2.9 m

D = OA + AB

= 1.5 + 2.9

= 4.4 M

Distance of key from edge = 4.4 m .

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