If the refractive index of a medium is 1.33, what is the maximum angle of incidence for total internal reflection when light travels to air?
Practice Questions
1 question
Q1
If the refractive index of a medium is 1.33, what is the maximum angle of incidence for total internal reflection when light travels to air?
41.8°
48.6°
53.1°
60.0°
The critical angle θc can be calculated as θc = sin^(-1)(n2/n1) = sin^(-1)(1.00/1.33) ≈ 48.6°.
Questions & Step-by-step Solutions
1 item
Q
Q: If the refractive index of a medium is 1.33, what is the maximum angle of incidence for total internal reflection when light travels to air?
Solution: The critical angle θc can be calculated as θc = sin^(-1)(n2/n1) = sin^(-1)(1.00/1.33) ≈ 48.6°.
Steps: 6
Step 1: Understand that the refractive index (n) of a medium tells us how much light bends when it enters that medium. Here, n1 (for the medium) is 1.33 and n2 (for air) is approximately 1.00.
Step 2: Recall the formula for the critical angle (θc) for total internal reflection, which is θc = sin^(-1)(n2/n1).
Step 3: Substitute the values into the formula: θc = sin^(-1)(1.00/1.33).
Step 4: Calculate the value of 1.00 divided by 1.33, which is approximately 0.7519.
Step 5: Use a calculator to find the inverse sine (sin^(-1)) of 0.7519. This gives you the critical angle θc.
Step 6: The result is approximately 48.6 degrees, which is the maximum angle of incidence for total internal reflection when light travels from the medium to air.
