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PhysicsMediumMCQ2021 · 18 Mar Shift 2

Q17.Three rays of light, namely red (R), green (G) and blue (B) are incident on the face PQ of a right angled prism PQR as shown in figure. The refractive indices of the material of the prism for red, green and blue wavelength are 1. 27, 1. 42 and 1. 49 respectively. The colour of the ray(s) emerging out of the face PR is : (1) green (2) red (3) blue and green (4) blue

What This Question Tests

This question requires calculating the critical angle for different colors of light within a prism and determining which colors undergo total internal reflection and thus do not emerge from a specific face.

Concepts Tested

Total internal reflectionCritical angleSnell's lawDispersion

Formulas Used

sin C = n_air/n_medium

📚 NCERT Sections This Tests

9.21At What Angle Should A Ray Of Light Be Incident On The Face Of A Prism

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9.21 At what angle should a ray of light be incident on the face of a prism of refracting angle 60° so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is 1.524.

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

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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

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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.