Consider a ray at the interface air and glass

[Gamma]

Consider a ray at the interface air and glass (n = 1.5). We know that incident angle i and refracted angle r are related by sin i = 1.5 sin r

Can we some how show the relationship between i and r is r = m i + c (straight line)?

If you look at the following data, it follows the Snell's law and r = 0.6 i + 1.

i r
10 7
20 13
30 19
40 25
50 31

 

[lightarrow]

Consider a ray at the interface air and glass (n = 1.5). We know that incident angle i and refracted angle r are related by sin i = 1.5 sin r

Can we some how show the relationship between i and r is r = m i + c (straight line)?

If you look at the following data, it follows the Snell's law and r = 0.6 i + 1.

i r
10 7
20 13
30 19
40 25
50 31

Hoping this is not an homework, write r as function of i using Snell's law, then develop the functions sin and arcsin to the appropriate order (in radians), considering that 50° are less than 1 radian.

 

[sir-pinski]

Yes, we can show the relationship between the incident angle i and refracted angle r as a straight line. This is because the equation sin i = 1.5 sin r follows the general form of a straight line equation, y = mx + c, where y = sin i, m = 1.5 and x = sin r. Therefore, we can rearrange the equation to get sin r = (1/1.5)sin i or r = (1/1.5)i + c, where c is a constant. This means that as the incident angle i increases, the refracted angle r will also increase at a constant rate of 1/1.5.

The data provided also supports this relationship, as the values of r increase in a linear manner as i increases. This is in accordance with Snell's law, which states that the ratio of the sine of the incident and refracted angles is constant for a given pair of media.

In conclusion, the relationship between the incident angle and refracted angle at the air-glass interface can be represented as a straight line with a slope of 1/1.5 and a y-intercept of c. This is a useful way to visualize and understand the behavior of light at this interface.