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
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