Question

In searching the bottom of a pool at night, a watchman shinesa

narrow beam of light from his flashlight, 1.3 m above the
waterlevel, onto the surface of the water at a point 2.7 m from the
edgeof the pool (Figure 23-50). Where does the spot of light hit
thebottom of the pool, measured from the wall beneath his foot, if
thepool is 2.1 m deep?

Answer 1

The distance of spot of light from his feet equals 3.425 meters.

Step-by-step explanation:

The situation is represented in the attached figure below

The angle of incidence is computed as

`\theta _i=tan^(-1)((1.3)/(2.7))\\\\\therefore \theta _i=25.71^(o)`

Now by Snell's law we have

`n_(i)sin(\theta _i)=n_(r)sin(\theta _r)`

where

`n_(i),n_(r)` are the refractive indices of the incident and the refracting medium respectively

`\theta _i,\theta _r` are the angle of incidence and the angle of refraction respectively

Thus using the Snell's relation we have

`1.0* sin(25.71)=1.33* sin(\theta _r)\\\\\therefore sin(\theta _r)=(sin(25.71)/(1.33)=0.326\\\\\therefore \theta _r=sin^(-1)(0.326)=19.04^(o)`

from the attached figure we can see

`tan(\theta _r)=(L_(2))/(H)=(L_(2))/(2.1)\\\\\therefore L_(2)=2.1* tan(19.04)=0.725m`

Thus distance of spot on the pool bed from his feet equals
`2.7+0.725=3.425m`

Answer 2

The spot of light hits the bottom of the pool at 4.634 m from the wall beneath the watchman's feet.

Step-by-step explanation:

Use the diagram attached below this answer to see the notation we will use.

For this case, we're trying to find x and we have:

h=1.3 m

b=2.1 m

a=2.7 m

We also know Snell's law for refraction:

`n_(1) sin\theta_(1)=n_(2)sin\theta_(2)`

n is the refractive index of each substance (in this case, air and water), which are:

`n_(air)=1`

`n_(water)=1.33`

Triangle theory says that
`\theta_(1)=\beta` and:

`tan\beta=(a)/(h)`

`\beta=arc tan((a)/(h))=arctan((2.7m)/(1.3m))=64.29`

Using Snell's law:

`\theta_(2)=arcsin(\frac{n_(1)sin\theta{1}}{n_(2)})=arcsin((1sin(64.29))/(1.33))=42.644`

Using triangle theory:

`tan\theta_(2)=((x-a))/(b)`

`x=b*tan\theta_(2)+a=2.1m*tan(42.644)+2.7m=4.634m`