Q51P - University Physics with Modern Physics

Question

When the sun is either rising or setting and appears to be just on the horizon, it is in fact below the horizon. The explanation for this seeming paradox is that light from the sun bends slightly when entering the earth’s atmosphere, as shown in Fig. Since our perception is based on the idea that light travels in straight lines, we perceive the light to be coming from an apparent position that is an angleabove the sun’s true position.

(a) Make the simplifying assumptions that the atmosphere has uniform density, and hence uniform index of refraction n, and extends to a height h above the earth’s surface, at which point it abruptly stops. Show that the angle  is given by

$\delta =arc\sin \left(\frac{nR}{R+h}\right)-arc\sin \left(\frac{R}{R+h}\right)$

where R = 6378 km is the radius of the earth. 

(b) Calculate $\delta$ using n = 1.0003 and h = 20 km. How does this compare to the angular radius of the sun, which is about one-quarter of a degree? (In actuality a light ray from the sun bends gradually, not abruptly, since the density and refractive index of the atmosphere change gradually with altitude.)

                                                        

Step-by-Step Solution

Answer

a) Proven in step 1

b) $\delta =0.{23}^o$ and it is almost the same for the angular radius of the sun.

Step 1: Prove δ = a r c sin n R R + h - a r c sin R R + h

 

The beam of the light incident at an angle ${\theta }_a$ and bends to refract at the angle ${\theta }_b$. Angle ${\theta }_b$ is less than angle ${\theta }_a$, and the difference between both of them is 

                                            $\delta ={\theta }_a-{\theta }_{ b}$               

Now, let us find an expression for each angle. From Figure, in the triangle with the sides R and R+ h, the sin of the angle ${\theta }_b$ is given by

                                                   $\sin {\theta }_b=\frac{R}{R+h}$      

Where R is the radius of the Earth while h is the height of the atmosphere 

Now, we get the refractive angle by

                                                ${\theta }_b=arc\sin \left(\frac{R}{R+h}\right)$   

The refractive index of an optical material is expressed by n and represents the speed of light in the vacuum divided by the speed of light in the material. Snell's law is given by an equation in the form 

                                                 $n_a\sin {\theta }_a=n_0\sin {\theta }_b$                               (1) 

Where $n_a$ for the vacuum equals 1 and $n_b$ for the air could be replaced by n. So, equation (1) will be

                       $n_a\sin {\theta }_a=n_b\sin {\theta }_b=n\left(\frac{R}{R+h}\right)$                        (2)

Hence, the angle ${\theta }_a$ is 

                                       ${\theta }_a=arcs\mathrm{in} \left(\frac{nR}{R+h}\right)$                

Now, we can get the angle $\delta$ by

                                        $$\begin{gathered} \delta ={\theta }_a-{\theta }_b \\ \delta =arc\sin \left(\frac{nR}{R+h}\right)-arc\sin \left(\frac{R}{R+h}\right) \end{gathered}$$   

Hence, we prove the requirements.

Step 2: Calculate δ

We are given R= 6378 km, n=1.0003 and the height is h = 20 km. Now we use the final equation by plugging the values for n, R, and h to get $\delta$

                                $\begin{array}{r}\delta =\arcsin \left(\frac{nR}{R+h}\right)-\arcsin \left(\frac{R}{R+h}\right) \\ =\arcsin \left(\frac{\left(1.0003\right)\left(6378\times {10}^3\mathrm{m}\right)}{6378\times {10}^8\mathrm{m}+20\times {10}^{\mathrm{s}}\mathrm{m}}\right)-\arcsin \left(\frac{6378\times {10}^3\mathrm{m}}{6378\times {10}^3\mathrm{m}+20\times {10}^8\mathrm{m}}\right)\\ ={0.23}^{\circ } \end{array}$

 

The angular radius of the sun is about one-quarter of a degree, so our result is almost the same for the angular radius of the sun.