Q2E - University Physics with Modern Physics

Question

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 380 nm (violet) to 750 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. What are the ranges of (a) the wavelength, (b) the frequency, and (c) the speed of the light just as it approaches the retina within the vitreous humor?

Step-by-Step Solution

Answer

1. The range of the wavelength is from 284 nm to 560 nm .
2. The range of the frequency is from  $4.00\times {10}^{14}\mathrm{Hz}$ to $7.89\times {10}^{14}\mathrm{Hz}$ .
3. The speed of the light as it approaches the retina is $2.24\times {10}^8\mathrm{m}/\mathrm{s}$.

Step 1: (a) Determination of wavelength.

The speed changes, so the wavelength also changes accordingly.

The refractive index n = 1.34 . So the shift in wavelength is,

For vitreous humour,

$$\begin{gathered} {\lambda }_v=\frac{{\lambda }_{0,v}}{n} \\ =\frac{380 nm}{1.34} \\ =284 nm \end{gathered}$$

For retina,

$$\begin{gathered} {\lambda }_v=\frac{{\lambda }_{0,r}}{n} \\ =\frac{750 nm}{1.34} \\ =560 nm \end{gathered}$$

Thus, the range of wavelength is from 284 nm to 560 nm 

Step 2: (b) Determination of the frequency.

The frequency expression is,

$f=\frac{v}{\lambda }$

The frequency in air with $\mathrm{v}=\mathrm{c}=3\times {10}^8\mathrm{m}/\mathrm{s}$,

$$\begin{gathered} f_r=\frac{c}{{\lambda }_r} \\ =\frac{3.00\times {10}^{8}m/s}{750\times {10}^{-9}m} \\ =4.00\times {10}^{14}Hz \end{gathered}$$

The frequency for vitreous humour,

$$\begin{gathered} f_r=\frac{c}{{\lambda }_v} \\ =\frac{3.00\times {10}^{8}m/s}{750\times {10}^{-9}m} \\ =7.89\times {10}^{14}Hz \end{gathered}$$

Thus, the frequency ranges from $4.00\times {10}^{14}Hz$ to $7.89\times {10}^{14}Hz$ .

Step 3: (c) Determination of the speed of light as it approaches the retina.

The expression for the speed is,

 $v=\frac{c}{n}$ 

Substitute all the values in the above equation,

  $$\begin{gathered} v=\frac{3.00\times {10}^{8}m/s}{1.34} \\ =2.24\times {10}^{8}m/s \end{gathered}$$

Thus, the speed with which light reaches the retina is $2.24\times {10}^{8}m/s$