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hellp me A certain species of fish is startled whenever it sees another animal 20 cm away, quickly running away. A fish of this species is in a cubic aquarium 5 cm away from the inside of the glass, while a person approaches to observe it more closely. The glass has n = 1.55 and is 3 cm thick. Assume that the person moves facing the fish and that the refractive index of water is 1.33. There is a first image of the person produced by refraction in the glass, and this image in turn functions as an object to produce a second image by refraction in water. (a) When the fish is startled and runs away, how far from the water/glass interface should the first image be? (b) How far from the outside surface of the aquarium is the person when this happens?

Sagot :

The medium of water, glass and air, form a composite, that result in

successive refraction of the light from the person.

Responses (approximate values):

(a) 23.25 cm

(b) 14.3 cm

How is the distance of the image found?

μ₁ = Refractive index of water = 1.33

Depth of fish, t₁ = 5 cm

μ₂ = Refractive index of glass = 1.55

Thickness of glass, t₂ = 3 cm

μ₃ = Refractive index of air = 1.00

t₃ = Location of the person outside the glass surface

(a) Refraction through multiple medium is given as follows;

[tex]\dfrac{\mu_2}{\mu_1} = \mathbf{ \dfrac{Image \ distance}{Object \ distance}}[/tex]

First image = Object distance from water/glass interface

Relative refractive index of glass and air = 1.55

Apparent depth of the of the first image from the water/glass interface is 20 cm - 5 cm = 15 cm

When the fish is startled, we have;

[tex]15 = \mathbf{\dfrac{t_2}{\mu_2} + \dfrac{t_1}{\mu_3}}[/tex]

Which gives;

[tex]15 = \dfrac{3}{1.55} + \dfrac{t_1}{1}[/tex]

Therefore;

3 + 1.55·t₁ = 1.55 × 15 = 23.25

  • First image distance from the water/glass interface = 23.25  cm

(b) The distance from the outside of the aquarium is found as follows;

[tex]Apparent \ Depth = \mathbf{\dfrac{t_1}{\mu_1} + \dfrac{t_2}{\mu_2} + \dfrac{t_1}{\mu_3}}[/tex]

Therefore;

[tex]20 = \mathbf{\dfrac{3}{1.55} + \dfrac{5}{1.33} + \dfrac{t_1}{1}}[/tex]

Which gives;

t₁ ≈ 14.3 cm

  • The distance of the person from the outside surface of the aquarium when it happens, t₁ ≈ 14.3 cm

Learn more about refractive index here:

https://brainly.com/question/24448445

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