Total Internal Reflection and Critical Angle - UNSOLVED PRACTICE SET
Chapter: Ray Optics and Optical Instruments | Topic: Total Internal Reflection and Critical Angle
TOTAL INTERNAL REFLECTION AND CRITICAL ANGLE - UNSOLVED PRACTICE SET
Topic: Total Internal Reflection and Critical Angle
Multiple Choice Questions
Q1. Total internal reflection occurs when light travels from:
- Rarer to denser medium
- Denser to rarer medium
- Any medium to any other medium
- Vacuum to any medium
Q2. The critical angle is the angle of incidence in the denser medium for which the angle of refraction is:
- 0°
- 45°
- 90°
- 180°
Q3. The critical angle for a pair of media is related to their refractive indices by:
- sin C = n₁/n₂ where n₁ > n₂
- sin C = n₂/n₁ where n₁ > n₂
- sin C = n₁ × n₂
- sin C = n₁ + n₂
Q4. Mirage in deserts is caused by:
- Refraction only
- Total internal reflection
- Reflection from sand
- Diffraction
Q5. Optical fibers work on the principle of:
- Refraction
- Total internal reflection
- Dispersion
- Interference
Q6. For total internal reflection to occur, the angle of incidence must be:
- Less than the critical angle
- Equal to the critical angle
- Greater than the critical angle
- Equal to 90°
Short Answer Questions
Q7. Define critical angle. What are the two essential conditions for total internal reflection to occur?
Q8. Explain why total internal reflection cannot occur when light travels from air to water.
Q9. What is an optical fiber? Draw a diagram showing how light travels through it using total internal reflection.
Q10. Why does a diamond sparkle? Explain the role of total internal reflection and the high refractive index of diamond.
Q11. Explain the formation of a mirage on a hot summer day using the concept of total internal reflection.
Q12. Calculate the critical angle for a glass-air interface if the refractive index of glass is 1.5.
Long Answer Questions
Q13. Derive the expression for critical angle in terms of refractive indices of two media. Explain the conditions under which total internal reflection occurs.
Q14. Explain the working of an optical fiber using the principle of total internal reflection. Discuss the structure of an optical fiber and its applications in communication.
Q15. With the help of a ray diagram, explain the formation of a mirage in a desert. How does the variation of refractive index of air with temperature contribute to this phenomenon?
Numerical & Application-Based Problems
Q16. The refractive index of water is 4/3 and that of glass is 3/2.
(a) Calculate the critical angle for a water-glass interface (light going from glass to water).
(b) Calculate the critical angle for a glass-air interface.
(c) Calculate the critical angle for a water-air interface.
(d) Arrange these critical angles in increasing order and explain the physical significance.
Q17. An optical fiber has a core of refractive index 1.60 and a cladding of refractive index 1.48.
(a) Calculate the critical angle for the core-cladding interface.
(b) Calculate the maximum angle of acceptance (acceptance angle) for light entering the fiber from air.
(c) Explain what happens if light enters the fiber at an angle greater than the acceptance angle.
Q18. In your school's science exhibition, a student demonstrates total internal reflection using a semicircular glass block of refractive index 1.5.
(a) Calculate the critical angle for the glass-air interface.
(b) The student directs a laser beam onto the flat face of the block such that it hits the curved surface at an angle of 50° to the normal. Predict whether total internal reflection will occur and explain your reasoning.
(c) The student then places the block in water (refractive index 4/3). Calculate the new critical angle for the glass-water interface.
(d) A classmate asks why diamonds sparkle more than glass cut in the same shape. Explain using the concepts of critical angle and total internal reflection, calculating the critical angles for both diamond (n = 2.42) and glass (n = 1.5).
(e) Fiber optic cables are being laid across India for high-speed internet. Explain why the core of the fiber must have a higher refractive index than the cladding, and what would happen if they had the same refractive index.