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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?

User Ewald
by
4.5k points

2 Answers

5 votes

Answer:

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

In searching the bottom of a pool at night, a watchman shinesa narrow beam of light-example-1
User Anubrij Chandra
by
5.1k points
4 votes

Answer:

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

In searching the bottom of a pool at night, a watchman shinesa narrow beam of light-example-1
User PatrikJ
by
5.4k points