A swimmer is treading water with their head above the surface of a pool and sees a penny at the bottom of the pool 5.0 mm below. How deep does the coin appear to be? (Index of refraction of water = 1.33) [Conceptual note: Does the coin appear to be shallower or deeper?]

Respuesta :

Answer:

The  apparent depth is  [tex]D' = 0.00376 \ m[/tex]

Explanation:

From the question we are told that

     The  depth of the water is  [tex]D = 5.0 \ mm = 5.0 *10^{-3} \ m[/tex]

      The  refractive index of water is  [tex]n = 1.33[/tex]

     

Generally the apparent depth of the coin is mathematically represented as

          [tex]D' = D * [\frac{ n_a}{n} ][/tex]

Here  [tex]n_a[/tex]  is the refractive index of  air the value is  [tex]n_a = 1[/tex]

So

        [tex]D' = 5.0 *10^{-3} * [\frac{1}{1.33} ][/tex]

        [tex]D' = 0.00376 \ m[/tex]

The apparent depth will be 0.00376 m.

What is an index of refraction?

The index of refraction of a substance also known as the refraction index is a dimensionless quantity that specifies how quickly light passes through it in optics.

d is the depth of the water =5.0 mm =5.0 ×10⁻³

n is the refractive index of water =1.33

[tex]\rm n_a[/tex] is the refractive index of wire=1

The apparent depth of the coin is given as;

[tex]\rm D'=D \times \frac{n_a}{n} \\\\ \rm D'=5.0 \times 10^{-3} \times \frac{1}{1.33} \\\\ \rm D'=0.00376 \ m[/tex]

Hence the apparent depth will be 0.00376 m.

To learn more about the index of refraction refer to the link;

https://brainly.com/question/23750645