A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be 40°. What is the refractive index of the material of the prism? The refracting angle of the prism is 60°. If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light.
1. Given Data
- Angle of Prism (A): 60°
- Minimum Deviation in Air (δm): 40°
- Refractive Index of Water (nw): 1.33
2. Calculating Refractive Index of Prism (ng)
Using the Prism Formula:
n = sin[(A + δm)/2] / sin(A/2)
Substituting the values:
ng = sin((60° + 40°)/2) / sin(60°/2)
ng = sin(50°) / sin(30°)
Using standard trigonometric values (sin 50° ≈ 0.766, sin 30° = 0.5):
ng = 0.766 / 0.5
ng = 1.532
3. Calculating New Angle of Deviation in Water
When the prism is immersed in water, the refractive index of the prism relative to water (n') is:
n' = ng / nw
n' = 1.532 / 1.33
n' ≈ 1.152
Now, applying the Prism Formula for the new deviation (δ'm):
n' = sin[(A + δ'm)/2] / sin(A/2)
1.152 = sin[(60° + δ'm)/2] / sin(30°)
1.152 = sin[(60° + δ'm)/2] / 0.5
Rearranging:
sin[(60° + δ'm)/2] = 1.152 × 0.5
sin[(60° + δ'm)/2] = 0.576
taking inverse sine (sin-1(0.576) ≈ 35.16°):
(60° + δ'm)/2 = 35.16°
60° + δ'm = 70.32°
δ'm = 70.32° - 60°
δ'm ≈ 10.32°
Conclusion: Placing the prism in water reduces the refractive difference between the prism and its surroundings, significantly lowering the angle of deviation from 40° to approximately 10.32°.
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