Chapter 34: Q 52 (page 992)

It’s night time, and you’ve dropped your goggles into a 3.0-m-deep swimming pool. If you hold a laser pointer 1.0 m above the edge of the pool, you can illuminate the goggles if the laser beam enters the water 2.0 m from the edge. How far are the goggles from the edge of the pool?

Short Answer

Expert verified

The distance of the goggles from the edge of the pool is 4.73 m

Step by step solution

Step 1. Given information is :Depth of the swimming pool = 3 mLaser pointer is 1 m above the edge of the poolLaser illuminates the goggles if it enters the water 2.0 m from the edgeRefractive index of air n1 = 1Refractive index of Water n2 = 1.33

We need to find out the distance of the goggles from the edge of the pool.

Step 2. Applying Snell's Law

Let $ϕ$ be the angle made by the laser ray with the surface of the pool as shown below.

Therefore,

$\tan ϕ = \frac{1.0 m}{2.0 m} ϕ = \tan^{- 1} (\frac{1}{2}) = 26.56 °$

Then angle of incidence $θ_{1}$ for the laser ray is :

localid="1648326367714" $θ_{1} = 90 ° - 26.56 ° θ_{1} = 63.44 °$

Using Snell's Law,

localid="1648326669186" $n_{1} \sin θ_{1} = n_{2} \sin θ_{2} \sin θ_{2} = \frac{n_{1}}{n_{2}} \sin θ_{1} θ_{2} = \sin^{- 1} (\frac{n_{1}}{n_{2}} \sin θ_{1}) θ_{2} = \sin^{- 1} (\frac{1.00}{1.33} \sin (63.44) °) θ_{2} = 42.26 °$

The distance x can be calculated by :

$\tan θ_{2} = \frac{x}{3.0 m} x = (3.0 m) \times \tan (42.26) ° x = 2.73 m T h e d i s \tan c e b e t w e e n e d g e o f t h e p o o l a n d g o g g l e s i s (2 + 2.73) m = 4.73 m$