The vitreous humor, a transparent, gelatinous fluid that fills most of the eyeball, has an index of refraction of 1.34. Visible light ranges in wavelength from 400 nm (violet) to 700 nm (red), as measured in air. This light travels through the vitreous humor and strikes the rods and cones at the surface of the retina.

A: What are the ranges of the wavelength of the light just as it approaches the retina within the vitreous humor?
Answer in the order indicated. Separate your answers with a comma.

λmin,λmax=

B: What are the ranges of the frequency of the light just as it approaches the retina within the vitreous humor?
Answer in the order indicated. Separate your answers with a comma.

fmin,fmax=

C:What is the speed of the light just as it approaches the retina within the vitreous humor?

v=

When the light passes through a material medium to another the electrons of the atoms exist, as a formed oscillation, these electors re-emit the radiation with the same frequency, process is resonant, but by the index of refraction we know that the speed of light in the medium has changed, therefore the wavelength must be changed to meet the relationship

In a vacuum c = λ₀ f

In a material v = λ f

n = c / v = λ₀/λ

λ = λ₀ / n

.a) Let's calculate the new wavelengths in the vitreous humor

With n = 1.34

λ₁ = 400 / 1.34

λ₁ = 298.5 nm

λ₂ = 700 / 1.34

λ₂ = 522.4 nm

b) as the frequency does not change when changing the medium we can use the values for the vacuum or air

c = λ₀ f

f₁ = c /λ₀

f₁ = 3 10⁸/400 10⁻⁹

f₁ = 7.5 10¹⁴ Hz

f₂ = 3 10⁸/700 10⁻⁹

f₂ = 4.29 10¹⁴ Hz

c) the speed of light in this material medium

v = λ f

v = 298.5 10⁻⁹ 7.5 10¹⁴

v = 2,238 10⁸ m / s

n = c / v

v = c / n

v = 3 10⁸ / 1.34

v = 2,238 10⁸ m / s

batolisis batolisis · Answer 2 · 2022-03-25T17:44:31+01:00

The answers to the question are as listed below :

A) The ranges of the wavelength of the light as it approaches the retina are ; λmin = 298.5 nm, λmax = 522.4 nm

B) The ranges of the frequency of light are as follows : fmin = 7.5 * 10¹⁴ Hz, fmax = 4.29 * 10¹⁴ Hz ,

C) The speed of light as it approaches the retina : v = 2238 * 10⁸ m / s

A) Determine the ranges of wavelength of the light

applying the formula

λmin = λ₀ / n

Given that n = 1.34 , λ₀ = 400

λmin = 400 / 1.34

= 298.5 nm

λmax = 700 / 1.34

= 522.4 nm

B) Determine the ranges of frequency

Given that frequency does not change with medium we will apply the frequency of a vacuum

Given that C = λ₀ f

fmin = c / λ₀

therefore fmin = 3 * 10⁸ /400 * 10⁻⁹

fmin = 7.5 * 10¹⁴ Hz

Also

fmax = 3* 10⁸ / 700 10⁻⁹

fmax = 4.29 * 10¹⁴ Hz

C) Determine the speed of the light

we will apply the formula below

v = λmin * fmin

therefore: v = 298.5 * 10⁻⁹ * 7.5 * 10¹⁴

v = 2,238 * 10⁸ m / s

Hence we can conclude that the answers to your questions are as listed above

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