RankLab
Back to Questions
PhysicsMediumNumerical2014 · 09 Apr Online

Q23.A diver looking up through the water sees the outside world contained in a circular horizon. The refractive index of water is 4 , and the diver's eyes are 15 cm below the surface of the water. Then the radius of the circle 3 is : (1) 15×√7 3 cm (2) 15×3√7 cm (3) 15 × 3√7 cm (4) 15 × 3 × √5 cm

What This Question Tests

This question applies the concept of total internal reflection to determine the radius of the circular horizon a diver sees, using the critical angle and basic geometry.

Concepts Tested

Critical angleSnell's LawTotal Internal ReflectionGeometry

Formulas Used

sin θ_c = n_air/n_water

R = h tan θ_c

📚 NCERT Sections This Tests

9.5A Small Bulb Is Placed At The Bottom Of A Tank Containing Water To A

Physics Class 12 · Chapter 9

82% match

9.5 A small bulb is placed at the bottom of a tank containing water to a depth of 80cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.)

9.3A Tank Is Filled With Water To A Height Of 12.5 Cm. The Apparent

Physics Class 12 · Chapter 9

79% match

9.3 A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?

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.