a ray in oil (n = 1.52) reaches a boundary with water (n = 1.33) at 55.9 deg. does it reflect internally or…
a ray in oil (n = 1.52) reaches a boundary with water (n = 1.33) at 55.9 deg. does it reflect internally or refract into the water? enter 0 for reflect, and 1 for refract. (oil n = 1.52, water n = 1.33)
Answer
Explanation:
Step1: Calculate the critical angle
The formula for the critical angle ( \theta_c) is ( \sin\theta_c=\frac{n_2}{n_1}), where (n_1 = 1.52) (refractive index of oil) and (n_2=1.33) (refractive index of water). [ \sin\theta_c=\frac{1.33}{1.52} ] [ \theta_c=\sin^{- 1}\left(\frac{1.33}{1.52}\right)\approx61.0^{\circ} ]
Step2: Compare the incident angle with the critical angle
The incident angle (\theta = 55.9^{\circ}) Since (\theta=55.9^{\circ}<\theta_c = 61.0^{\circ})
Answer:
1