Question

The refractive index of vitreous humor in our eye ball is 1.34. Visible light of wavelength ranges from 402 nm to 670 nm as measured in air. The wavelength range just approaches retina within vitreous humor is: A) 340 nm to 540 nm B) 280 nm to 480 nm c) 300 nm to 500 nm D) 320 nm to 520 nm​

Answer 1

Answer:

yes

Explanation:

I hope it will be help you with

Answer 2

Answer:

option C) 300 nm to 500 nm

Explanation:

Given :

• The refractive index of vitreous humor in our eye ball is 1.34.
• Visible light of wavelength ranges from 402 nm to 670 nm as measured in air.

To find :

the wavelength range just approaches retina within vitreous humor

Solution :

The relation between speed of light, wavelength and frequency is :

In vacuum, c = nλ₀  

In any medium, the speed of the light changes.

v = nλ  

We also know that the refractive index is the ratio of the speed of the light in vacuum to that of the medium.

$\longmapsto \sf \mu=\dfrac{c}{v} \\\\ \longmapsto \sf \mu=\dfrac{\lambda _{o}}{\lambda} \\\\ \implies \boxed{\sf \lambda=\dfrac{\lambda _{o}}{\mu}}$

Given, the wavelength of the visible light ranges from 402 nm to 670 nm as measured in air.

 Let λ₁ = 402 nm

  λ₂ = 670 nm

Let's calculate the required wavelength range.

$\sf \lambda _1 '=\dfrac{402 \ nm}{1.34} \\\\ \sf \lambda _1 '=\dfrac{402}{134} \times 100 \: nm \\\\ \sf \lambda _1 '=300 \: nm$

___________________

$\sf \lambda _2 '=\dfrac{670 \ nm}{1.34} \\\\ \sf \lambda _2 '=\dfrac{670}{134} \times 100 \: nm \\\\ \sf \lambda _2 '=500 \: nm$

Therefore, the required wavelength range is 300 nm to 500 nm