Q25.Unpolarised light is incident on the boundary between two dielectric media, whose dielectric constants are 2. 8 (medium – 1) and 6. 8 (medium – 2), respectively. To satisfy the condition, so that the reflected and refracted 1 rays are perpendicular to each other, the angle of incidence should be tan−1 (1 + 10θ ) 2 , the value of θ is ______. (Given for dielectric media, μr = 1 )
What This Question Tests
This question combines the concept of Brewster's law (reflected and refracted rays being perpendicular) with the relation between refractive index and dielectric constant to find the angle of incidence.
Concepts Tested
Formulas Used
tan ip = n2 / n1
n = √k
ip = tan⁻¹(n)
📚 NCERT Sections This Tests
9.4 — Figures 9.27(A) And (B) Show Refraction Of A Ray In Air Incident At 60°
Physics Class 12 · Chapter 9
9.4 Figures 9.27(a) and (b) show refraction of a ray in air incident at 60° with the normal to a glass-air and water-air interface, respectively. Predict the angle of refraction in glass when the angle of incidence in water is 45° with the normal to a water-glass interface [Fig. 9.27(c)]. FIGURE 9.27
9.6 — A Prism Is Made Of Glass Of Unknown Refractive Index. A Parallel
Physics Class 12 · Chapter 9
9.6 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.
9.17 — (A) Sin I¢C = 1.44/1.68 Which Gives I¢C = 59°. Total Internal Reflection
Physics Class 12 · Chapter 9
9.17 (a) sin i¢c = 1.44/1.68 which gives i¢c = 59°. Total internal reflection takes place when i > 59° or when r < rmax = 31°. Now, (sin i /sin r max max ) = 1.68 , which gives imax ~ 60°. Thus, all incident rays of angles in the range 0 < i < 60° will suffer total internal reflections in the pipe. (If the length of the pipe is finite, which it is in practice, there will be a lower limit on i determined by the ratio of the diameter to the length of the pipe.) (b) If there is no outer coating, i¢c = sin–1(1/1.68) = 36.5°. Now, i = 90° will have r = 36.5° and i¢ = 53.5° which is greater than i¢c. Thus, all incident rays (in the range 53.5° < i < 90°) will suffer total internal reflections.
📋 Question Details
- Chapter
- Ray Optics
- Topic
- Polarisation (Brewster's Law)
- Year
- 2023
- Shift
- 29 Jan Shift 2
- Q Number
- Q25
- Type
- Numerical
- NCERT Ref
- Class 12 Physics Ch 10: Wave Optics
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