Question

A layer of carbon dioxide, with index of refraction 1.00045 , rests on a block of ice, with index of refraction \(1.310 .\) A ray of light passes through the carbon dioxide at an angle of \(\varphi_{1}\) relative to the boundary between the materials and then passes through the ice at an angle of \(\varphi_{2}=72.06^{\circ}\) relative to the boundary. What is the value of \(\varphi_{1}\) ?

Short answer

Answer: The angle at which the light ray passes through the carbon dioxide layer is approximately 50.05°.

Step-by-step solution

Write Snell's law formula

Snell's law states that the ratio of the sines of the angles of incidence (in this case, \(\varphi_{1}\)) and refraction (in this case, \(\varphi_{2}\)) is equal to the inverse ratio of the indices of refraction of the two materials. The formula for Snell's law is given by: \(n_1 \cdot \sin{\varphi_1} = n_2 \cdot \sin{\varphi_2}\) where \(n_1\) and \(n_2\) are the indices of refraction of the carbon dioxide and ice layers, respectively.

Substitute the given values into Snell's law

Plug in the given values for the indices of refraction and the angle \(\varphi_2\) into Snell's law: \(1.00045 \cdot \sin{\varphi_1} = 1.310 \cdot \sin{72.06^{\circ}}\)

Solve for \(\varphi_1\)

To find the value of \(\varphi_1\), we need to isolate it in the equation. First, calculate the value of \(\sin{72.06^{\circ}}\) and multiply by the index of refraction of ice: \(1.00045 \cdot \sin{\varphi_1} = 1.310 \cdot 0.9511 \) Next, divide both sides of the equation by the index of refraction of carbon dioxide: \(\sin{\varphi_1} = \frac{1.310 \cdot 0.9511}{1.00045}\) Now, find the sine inverse of the result to obtain the value of \(\varphi_1\): \(\varphi_1 = \sin^{-1}(\frac{1.310 \cdot 0.9511}{1.00045})\) Calculating the sine inverse gives: \(\varphi_1 = 50.05^{\circ}\)

State the final answer

The value of the angle \(\varphi_1\) at which the light ray passes through the carbon dioxide layer is: \(\varphi_1 = 50.05^{\circ}\)