contestada

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?

Respuesta :

luisaugustosc

Answer:

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

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:

[tex]n_{1} sin\theta_{1}=n_{2}sin\theta_{2}[/tex]

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

[tex]n_{air}=1[/tex]

[tex]n_{water}=1.33[/tex]

Triangle theory says that [tex]\theta_{1}=\beta[/tex] and:

[tex]tan\beta=\frac{a}{h}[/tex]

[tex]\beta=arc tan(\frac{a}{h})=arctan(\frac{2.7m}{1.3m})=64.29[/tex]

Using Snell's law:

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

Using triangle theory:

[tex]tan\theta_{2}=\frac{(x-a)}{b}[/tex]

[tex]x=b*tan\theta_{2}+a=2.1m*tan(42.644)+2.7m=4.634m[/tex]

Ver imagen luisaugustosc
InesWalston

Answer:

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

Explanation:

The situation is represented in the attached figure below

The angle of incidence is computed as

[tex]\theta _i=tan^{-1}(\frac{1.3}{2.7})\\\\\therefore \theta _i=25.71^{o}[/tex]

Now by Snell's law we have

[tex]n_{i}sin(\theta _i)=n_{r}sin(\theta _r)[/tex]

where

[tex]n_{i},n_{r}[/tex] are the refractive indices of the incident and the refracting medium respectively

[tex]\theta _i,\theta _r[/tex] are the angle of incidence and the angle of refraction respectively

Thus using the Snell's relation we have

[tex]1.0\times sin(25.71)=1.33\times sin(\theta _r)\\\\\therefore sin(\theta _r)=\frac{sin(25.71}{1.33}=0.326\\\\\therefore \theta _r=sin^{-1}(0.326)=19.04^{o}[/tex]

from the attached figure we can see

[tex]tan(\theta _r)=\frac{L_{2}}{H}=\frac{L_{2}}{2.1}\\\\\therefore L_{2}=2.1\times tan(19.04)=0.725m[/tex]

Thus distance of spot on the pool bed from his feet equals [tex]2.7+0.725=3.425m[/tex]

Ver imagen InesWalston