NTA Abhyas JEE Main2020PhysicsRay OpticsPractice
Figure shows two spherical surface of radii R and 2 R separating three transparent media of refractive index μ = 4 , μ = 2 and μ = 1 . A ray of light travelling in a medium μ = 1 is incident on outer sphere tangentially. The net deviation suffered by light ray will be
Options
- A0 °
- B90 °
- C240 °
- D280 °
Correct answer
C. 240 °
Step-by-step solution
Path of light ray for minimum deviation will be sin 90 ° = 2 sin θ ⇒ θ = 30 ° Deviation = 60 ° There are 4 such deviations therefore total deviation = 240 ° .