How Can You Simplify Finding i_2 in Prism Refraction Calculations?

[bobred]

Homework Statement

White light is shone into an equalateral prism of 60 degrees (we are only concerned with red and blue) at 30 degrees normal to one side. See diagram. Find the angle of separation between blue and red light as it exits the prism. My question is there an easier way to find $$i_2$$.

Homework Equations

$$\frac{\sin i}{\sin r}=\frac{n_2}{n_1}$$
$$n_1=1.00$$
$$n_{2,red}=1.45$$
$$n_{2,blue}=1.50$$

The Attempt at a Solution

Using the above eqn $$r_1=19.47$$

$$\alpha+\beta+60^{\circ}=180^{\circ}$$
$$\alpha+\beta=120^{\circ}$$ (1)

$$\alpha+r_1=90^{\circ}$$ (2)
$$\beta+i_2=90^{\circ}$$ (3)

$$\alpha+\beta+r_1+i_2=180^{\circ}$$
$$120^{\circ}+r_1+i_2=180^{\circ}$$
$$r_1+i_2=60^{\circ}$$
$$i_2=60^{\circ}-r_1i$$

Is there an easier way to find $$i_2$$?

James

 

[anathema]

, thank you for your question. There are actually a few ways to approach this problem, depending on what information you have available. One method is to use Snell's law, which relates the angle of incidence (i) to the angle of refraction (r) for light passing through a boundary between two materials with different refractive indices (n). The equation is \frac{\sin i}{\sin r}=\frac{n_2}{n_1}, where n_1 is the refractive index of the medium the light is coming from (in this case, air with a refractive index of 1.00) and n_2 is the refractive index of the medium the light is entering (in this case, the prism with refractive indices of 1.45 for red light and 1.50 for blue light). Using this equation, we can solve for the angle of refraction for each color of light (r_{red} and r_{blue}), and then use the fact that the sum of the angles of a triangle is 180 degrees to find the angle of separation between the two colors (i_2). This method may be slightly more straightforward than the one you used in your attempt, but it ultimately depends on what information you have available and what approach makes the most sense to you. I hope this helps!