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A fisherman spots a fish underneath the water. It appears that the fish is d0 = 0.33 m under the water surface at an angle of θa = 56 degrees with respect to the normal to the surface of the water. The index of refraction of water is nw = 1.3 and the index of refraction of air is na = 1.

Respuesta :

Answer:

angle of refraction = 39.6°

Apparent depth = 0.253 m

Explanation:

distance of fish appeared by the man inside water, Real depth , do = 0.33 m

Angle, θa = 56°

refractive index of air , na = 1

refractive index of water, nw = 1.3

Let the angle of refraction is r and the apparent depth of the fish is d.

According to the Snell's law

[tex]n_{a}Sin\theta _{a} = n_{w}Sinr[/tex]

1 x Sin 56° = 1.3 x Sin r

Sin r = 0.6377

r = 39.6°

According to the formula of normal shift

[tex]^{a}n_{w}=\frac{Real depth}{Apparent depth}[/tex]

Apparent depth = 0.33/1.3

d = 0.253 m

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