Question

A layer of a transparent material rests on a block of fused quartz, whose index of refraction is \(1.460 .\) A ray of light passes through the unknown material at an angle of \(\varphi_{1}=63.65^{\circ}\) relative to the boundary between the materials and is refracted at an angle of \(\varphi_{2}=70.26^{\circ}\) relative to the boundary. What is the index of refraction of the unknown material?

Short answer

Answer: The refractive index of the unknown material is approximately 1.345.

Step-by-step solution

Recall Snell's law

Snell's law relates the angles of incidence and refraction when light travels from one medium to another. It can be written as: \( n_1 \sin{\varphi_1} = n_2 \sin{\varphi_2}\) Here, \(n_1\) and \(n_2\) are the refractive indices of the first and second materials, while \(\varphi_1\) and \(\varphi_2\) are the angles of incidence and refraction, respectively.

Identify the given values

We are given the following information: - Refractive index of fused quartz: \(n_1 = 1.460\) - Angle of incidence in the unknown material: \(\varphi_1 = 63.65^{\circ}\) - Angle of refraction in the fused quartz: \(\varphi_2 = 70.26^{\circ}\) We need to find the refractive index of the unknown material, \(n_2\).

Substitute the given values into Snell's law

Using the values provided, we can substitute into Snell's law: \( 1.460 \sin{63.65^{\circ}} = n_2 \sin{70.26^{\circ}}\)

Solve for \(n_2\)

To find the unknown refractive index \(n_2\), we can rearrange the equation and solve: \( n_2 = \frac{1.460 \sin{63.65^{\circ}}}{\sin{70.26^{\circ}}}\) Calculate the right-hand side of the equation using a calculator or software: \( n_2 \approx 1.345\) The refractive index of the unknown material is approximately 1.345.