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A flashlight is held at the edge of a swimming pool at a height of 1.5 m such that its beam makes an angle of 39 degrees with respect to the water's surface. The pool is 2.75 m deep and the index of refraction for air and water are 1 and 1.33, respectively. What is the horizontal distance from the edge of the pool to the bottom of the pool where the light strikes? Write your answer in meters.

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Answer:

The horizontal distance from the to the bottom of the pool is 3.831 m

Solution:

As per the question;

Height of the pool, H = 1.5 m

Depth of the pool, D =   2.75 m

Refractive index  of air, [tex]\mu_{a} = 1[/tex]

Refractive index  of air, [tex]\mu_{w} = 1.33[/tex]

Angle of the beam at the water surface, [tex]\theta = 39^{\circ}[/tex]

Now,

The angle of incidence is given by [tex]\angle i = 90^{\circ} - 39^{\circ} = 51^{\circ}[/tex]

Now, using Snell's law:

[tex]\mu_{a}sin(\angle i) = \mu_{w}sin\theta'[/tex]

[tex]1.sin(51^{\circ}) = 1.33sin\theta'[/tex]

[tex]0.777= 1.33sin\theta'[/tex]

[tex]theta' = 35.747^{\circ}[/tex]

Now,

From the normal, the horizontal image distance, d' is given by:

[tex]tan\theta' = \frac{d'}{d}[/tex]

[tex]d' = dtan(35.747) = 2.75\times 0.7198 = 1.979 m[/tex]

Now,

From the normal, the horizontal object distance, d'' is given by:

[tex]tani = \frac{d''}{H}[/tex]

[tex]d'' = Htan(51) = 1.5\times 1.2349 = 1.852 m[/tex]

Thus the horizontal distance is given by:

d' + d'' = 1.979 + 1.852 =3.831 m

ajeigbeibraheem

The horizontal distance from the edge of the pool to the bottom of the pool where light strikes is 3.83 m

What is Snell's Law of Refraction?

Snell's Law of refraction states that the proportion of the sine angle of incidence to the sine angle of refraction is constant for a wavelength of light.

It can be expressed by using the formula:

  • n₁sinθ₁ = n₂sinθ₂

where;

  • n₁ = refractive index of air = 1
  • n₂ = refractive index of water = 1.33
  • θ₁ = angle of incidence = 90° - 39° = 51°
  • θ₂ = angle of refraction = ???

n₁sinθ₁ = n₂sinθ₂

1(sin51°) = 1.33(sinθ₂)

sinθ₂ = 1(sin51°) ÷ 1.33

θ₂ = sin⁻¹(0.5843)
θ₂ = 35.75°

Taking the tangent of the horizontal distance to the height of the pool, we have:

[tex]\mathbf{ tan \ \theta = \dfrac{x}{2.75}}[/tex]

[tex]\mathbf{( tan\ 35.75 = \dfrac{x}{2.75})}[/tex]

x = tan 35.75 × 2.75

x = 0.7199 × 2.75

x = 1.979 m

Thus, from the edge of the pool, the horizontal distance is computed as:

[tex]\mathbf{ tan \ 51 = \dfrac{y}{1.2349}}[/tex]

y = 1.852 m

Finally, the horizontal distance from the edge of the pool to the bottom is:

= x + y

= 1.979 m + 1.852 m

≅ 3.83 m

Learn more about Snell's Law of Refraction here:

https://brainly.com/question/10112549

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