A fish is 4.7 cm from the front surface of a fish bowl of radius 21 cm. Where does the fish appear to be to someone in air viewing it from in front of the bowl? Do not forget the proper sign. (Give your answer in cm.)

cm

Where does the fish appear to be when it is 38.9 cm from the front surface of the bowl? (Give your answer in cm.)

cm

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ajeigbeibraheem ajeigbeibraheem · Answer 1 · 2020-12-04T03:45:12+01:00

Answer:

Explanation:

From the information given:

We can properly determine the distance where the fish appear in the air viewing it from in front of the bowl by using the formula:

[tex]\dfrac{n_i}{d_o}+\dfrac{n_2}{d_1}= \dfrac{n_2-n_1}{r}[/tex]

where;

[tex]n_1[/tex] = refractive index in the air; = 1.33 &

[tex]n_2[/tex] = refractive index in water. = 1

[tex]\dfrac{n_2}{d_i}= \dfrac{n_2-n_1}{r}-\dfrac{n_1}{d_o}[/tex]

[tex]\dfrac{1}{d_i}= \dfrac{1-1.33}{-21 \ cm}-\dfrac{1.33}{4.7\ cm}[/tex]

[tex]\dfrac{1}{d_i}= - 0.26726 \ cm[/tex]

[tex]d_i =\dfrac{1}{ - 0.26726 \ cm}[/tex]

[tex]\mathbf{d_i }[/tex] = - 3.74 cm

2)

To determine where the fish appear to be when it is 38.9 cm from the front surface of the bowl by using the formula:

[tex]\dfrac{n_2}{d_i}= \dfrac{n_2-n_1}{r}-\dfrac{n_1}{d_o}[/tex]

[tex]\dfrac{1}{d_i}= \dfrac{1-1.33}{-21 \ cm}-\dfrac{1.33}{38.9\ cm}[/tex]

[tex]\dfrac{1}{d_i}=- 0.0184759 \ cm[/tex]

[tex]d_i = \dfrac{1}{- 0.0184759 \ cm}[/tex]

[tex]\mathbf{d_i = }[/tex] -54.12 cm