A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm. 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.)
The area of the water surface through which light can emerge is approximately 2.61 m² (or 26,127 cm²).
Explanation
This problem is based on the concept of total internal reflection mentioned in Section 9.4 of the textbook. When light travels from water (optically denser medium, μ = 1.33) to air (rarer medium), it can only emerge out if the angle of incidence is less than the critical angle. Beyond the critical angle, total internal reflection occurs and light is reflected back into the water. The bulb acts as a point source, emitting light in all directions. Only rays striking the surface within a certain circular area (where angle of incidence < critical angle) can emerge. This forms a circular illuminated area on the water surface.
Solution Steps
Step 1: Find the critical angle for water-air interface For total internal reflection, sin(θc) = 1/μ = 1/1.33 = 0.7519 θc = sin⁻¹(0.7519) 48.75°
Step 2: Find the radius of the circular area From geometry, tan(θc) = r/d where r = radius of circular area, d = depth = 80 cm r = d × tan(θc) = 80 × tan(48.75°) = = 91.2 cm
Step 3: Calculate the area Area = πr² = π × (91.2)² = π × 8317.44 Area 26,127 cm² 2.61 m²