An experimental optical scientist is investigating the refractive properties of a newly synthesized polymer block.
She shines a narrow laser beam from air into the polymer block at various angles of incidence, θi\theta_iθi, and measures the corresponding angles of refraction, θr\theta_rθr. Her measurements are recorded in the table below:
| Angle of incidence, θi\theta_iθi | Angle of refraction, θr\theta_rθr | sinθi\sin \theta_isinθi | sinθr\sin \theta_rsinθr |
|---|---|---|---|
| 0∘0^\circ0∘ | 0∘0^\circ0∘ | 0.00 | 0.00 |
| 15∘15^\circ15∘ | 9∘9^\circ9∘ | 0.26 | 0.16 |
| 30∘30^\circ30∘ | 18∘18^\circ18∘ | 0.50 | [A] |
| 45∘45^\circ45∘ | 26∘26^\circ26∘ | 0.71 | [B] |
| 60∘60^\circ60∘ | 32∘32^\circ32∘ | 0.87 | 0.53 |
(i) Calculate the missing values [A] and [B] of sinθr\sin \theta_rsinθr from the table, giving your answers to 2 decimal places.
(ii) The scientist plots a graph with sinθi\sin \theta_isinθi on the vertical axis and sinθr\sin \theta_rsinθr on the horizontal axis. The linear line of best fit passes through the origin (0.00,0.00)(0.00, 0.00)(0.00,0.00) and the coordinate point (0.45,0.72)(0.45, 0.72)(0.45,0.72). Use these values to calculate the refractive index of the polymer block.
Suggest two reasons why finding the refractive index using a graphical line of best fit is superior to calculating it using a single pair of angles from the table.
