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PhysicsMediumMCQ2019 · 10 Apr Shift 1

Q24.A ray of light AO in vacuum is incident on a glass slab at angle 60° and refracted at angle 30° along OB as shown in the figure. The optical path length of light ray from A to B is: (1) 2√3 a + 2b (2) 2a + 2b (3) 2a + 2b (4) 2a + 2b3 √3

What This Question Tests

The problem requires applying Snell's law to find the refractive index and then calculating the optical path length by summing the product of refractive index and geometrical path length in each medium.

Concepts Tested

Snell's LawRefractionOptical path lengthTrigonometry

Formulas Used

μ₁ sinθ₁ = μ₂ sinθ₂

Optical path length = μ × geometrical path length

📚 NCERT Sections This Tests

9.4Figures 9.27(A) And (B) Show Refraction Of A Ray In Air Incident At 60°

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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

9.6A Prism Is Made Of Glass Of Unknown Refractive Index. A Parallel

Physics Class 12 · Chapter 9

78% match

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

78% match

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.