Ch 31: Geometric OpticsWorksheetSee all chapters
All Chapters
Ch 01: Units & Vectors
Ch 02: 1D Motion (Kinematics)
Ch 03: 2D Motion (Projectile Motion)
Ch 04: Intro to Forces (Dynamics)
Ch 05: Friction, Inclines, Systems
Ch 06: Centripetal Forces & Gravitation
Ch 07: Work & Energy
Ch 08: Conservation of Energy
Ch 09: Momentum & Impulse
Ch 10: Rotational Kinematics
Ch 11: Rotational Inertia & Energy
Ch 12: Torque & Rotational Dynamics
Ch 13: Rotational Equilibrium
Ch 14: Angular Momentum
Ch 15: Periodic Motion (NEW)
Ch 15: Periodic Motion (Oscillations)
Ch 16: Waves & Sound
Ch 17: Fluid Mechanics
Ch 18: Heat and Temperature
Ch 19: Kinetic Theory of Ideal Gasses
Ch 20: The First Law of Thermodynamics
Ch 21: The Second Law of Thermodynamics
Ch 22: Electric Force & Field; Gauss' Law
Ch 23: Electric Potential
Ch 24: Capacitors & Dielectrics
Ch 25: Resistors & DC Circuits
Ch 26: Magnetic Fields and Forces
Ch 27: Sources of Magnetic Field
Ch 28: Induction and Inductance
Ch 29: Alternating Current
Ch 30: Electromagnetic Waves
Ch 31: Geometric Optics
Ch 32: Wave Optics
Ch 34: Special Relativity
Ch 35: Particle-Wave Duality
Ch 36: Atomic Structure
Ch 37: Nuclear Physics
Ch 38: Quantum Mechanics

Solution: A person in the air above the water in a swimming pool looks straight down into the water (n = 1.33) at a diamond ring that lies on the bottom of the pool. If the image of the ring is 1.20 m below the

Problem

A person in the air above the water in a swimming pool looks straight down into the water (n = 1.33) at a diamond ring that lies on the bottom of the pool. If the image of the ring is 1.20 m below the surface of the water, what is the depth of the water (the distance from the surface of the water to the bottom of the pool)? Hint: Assume θ<<1, so sinθ=θ, and tanθ=θ, for a θ in radians.

(a) 0.60 m

(b) 0.90 m

(c) 1.20 m

(d) 1.60 m

(e) 2.40 m

(f) none of the above answers