Actual Depth and Apparent Depth

In summary, the problem involves finding the actual depth of a fish based on the apparent depth and angle of observation. By applying Snell's law for refraction, the angle of incidence can be found which leads to the calculation of the actual depth using basic trigonometric functions. The final solution is 3.2m for the actual depth of the fish.
  • #1
chibi_lenne
18
0
Okay, I need a bit of a jump start with this question, I know how to find the Apparent depth normally, but I've never done actual depth and I can't really figure it out (*stupid*)

11. Frederika is sitting in her fishing boat observing a rainbow trout swimming below the surface of the water. She guesses the apparent depth of the trout at 2.0m. She estimates that her eyes are about 1.0m above the water's surface, and that the angle at which she's observing the trout is 45degrees...

b)Calculate the actual depth of the trout.


please help ~_~
 
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  • #2
Ok I hope you know something about "what refractive index" does ?

If we are looking from air into water then the refractive index m is related as:

[itex]
m=\frac{realdepth}{apparentdepth}
[/itex]

First try to draw figure of above situation using correct laws for refraction and a good ray diagram will definitely help..show your work...help will follow..
 
  • #3
If your fuzzy on refraction you might want to take a look at this reference
If you read along, you will see they discuss depth perception.

I agree with Doc, to try and draw a complete diagram of the information given. It makes the analysis much easier.
 
  • #4
I decided to give my brain a bit of a rest on this problem and moved forward. I think I may have it, I'd still like to see if it's accurate or get help if it's completely wrong (which is very likely) Here's what I have so far:
Analysis:

n = 1.33 for water
n[2]= 1.00 for air

Actual depth= (sine{angle}i)(d)

tanZ = tan{angel}R = d/h, therefore d= (h)(tan{angle}R)

{Angle}R = (n)(sin{angle}i)/(n[2])

Solution:

sin{angle}R = (3.00)(0.071)/(1.00)
= 0.9404
{angle)R = 70.1*​

d = (3.0m)(tan70.1*)
= (3.0m)(2.762)
= 8.286m

Actual Depth = (sin{angle}i)(d)
= (sin45*)(8.286m)
= (0.7071)(8.286m)
= 5.859m​
Therefore the actual depth of the fish is 5.859m.

Hope this is right this problem is driving me nuts!
 

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  • #5
chibi_lenne said:
Actual depth= (sine{angle}i)(d)
According to your diagram, that should be: Actual depth = [itex]d/ \tan \theta_i[/itex]. Note that d = 2.0 m, since [itex]\theta_r[/itex] is 45 degrees.
 
  • #6
Doc Al said:
According to your diagram, that should be: Actual depth = [itex]d/ \tan \theta_i[/itex]. Note that d = 2.0 m, since [itex]\theta_r[/itex] is 45 degrees.

Now I'm confused...how is d = 2.0m? if I did the equation for actual depth with that as d, it would mean the actual depth is the same as the apparent depth?

Actual depth = (2.0m)/tan45*
= (2.0m)/(1.0)
= 2.0m​
 
Last edited by a moderator:
  • #7
chibi_lenne said:
Now I'm confused...how is d = 2.0m?
This can be deduced from the statement of the problem: The apparent depth is 2 m and the angle is 45 degrees.

if I did the equation for actual depth with that as d, it would mean the actual depth is the same as the apparent depth?
No.

Actual depth = (2.0m)/tan45*
= (2.0m)/(1.0)
= 2.0m​
No. Actual depth = [itex]d/ \tan \theta_i[/itex], not [itex]d/ \tan \theta_r[/itex]
 
  • #8
okay, so I have to find the angle of incidence...so how do I do that? x.x
 
  • #9
By applying Snell's law for refraction.
 
  • #10
duh *slaps forhead* alright I have to rearrange the formula, I feel stupid now. So it should be:

(ni)(sin{angle}i) = (nR)(sin{angle}R) so therefore

sin{angle}i = (ni)/(nR)(sin{angle}R)
?
 
  • #11
I believe you made an error in reaching your second equation.

Snell's law tells us:
[tex]n_1 \sin \theta_1 = n_2 \sin \theta_2[/tex]

So:
[tex]\sin \theta_1 = (n_2/n_1) \sin \theta_2[/tex]
 
  • #12
Doc Al said:
I believe you made an error in reaching your second equation.

Snell's law tells us:
[tex]n_1 \sin \theta_1 = n_2 \sin \theta_2[/tex]

So:
[tex]\sin \theta_1 = (n_2/n_1) \sin \theta_2[/tex]

ooo I see, (forgot to put in an extra set of () ) so it should be:

[tex] \sin \theta_i = (1.00/1.33) (0.8509)[/tex]
[tex]
= 0.6398 [/tex]
[tex] \theta_i = 39.6* [/tex]


??
 
  • #13
chibi_lenne said:
[tex] \sin \theta_i = (1.00/1.33) (0.8509)[/tex]
Where does the "0.8509" come from??
 
  • #14
Doc Al said:
Where does the "0.8509" come from??

[tex] 0.8509 = \sin45*[/tex]

Or at least that is what I was doing...
 
  • #15
Better double check that result!
 
  • #16
gah what am I doing wrong? x.x
 
  • #17
Ah, now I see what you did... You have your calculator set for radians, not degrees. :wink:
 
  • #18
*mutters something about incompitant calculators*

So it should be something like this?

[tex] \sin \theta_i = (n_i / n_R)\sin \theta_R [/tex]
[tex] = (1.00 / 1.33 )(\sin45*) [/tex]
[tex] = (0.7519)(0.7071) [/tex]
[tex] = 0.5317 [/tex]
[tex] \theta_i = 32.1* [/tex]

[tex] Actual Depth = (2.0m)/(\tan32.1*) [/tex]
[tex] = (2.0m)/(0.6277) [/tex]
[tex] = 3.18m [/tex]

Making the actual depth of the fish 3.2m??

I hope this is it, or close anyway. :cry:
 
  • #19
Looks good to me!
 
  • #20
Yay! Thanks so much! I really appreciate the help and for you putting up with my massive brain malfunction ^_^()
 

What is the difference between actual depth and apparent depth?

Actual depth refers to the true distance from the surface of an object to its deepest point, while apparent depth is the distance from the surface of an object to its perceived deepest point due to the bending of light.

How is actual depth and apparent depth related to refraction?

Refraction is the bending of light as it passes through different mediums, and it is the reason for the difference between actual depth and apparent depth. When light travels from a less dense medium to a more dense medium, such as from air to water, it slows down and bends, making the object appear closer and the apparent depth shorter.

What factors affect the difference between actual depth and apparent depth?

The main factor affecting this difference is the angle at which light enters and exits the object. The greater the angle of incidence, the greater the difference between actual depth and apparent depth. Another factor is the difference in the density of the two mediums that the light is traveling through.

How can we calculate the actual depth of an object?

To calculate the actual depth of an object, we need to know the refractive index of the medium the light is traveling through and the angle of incidence. Using Snell's law, we can calculate the angle of refraction and then use basic trigonometry to find the actual depth.

What are some real-life applications of understanding actual depth and apparent depth?

Understanding these concepts is essential in fields such as optics, underwater photography, and diving. It is also crucial in designing and constructing optical instruments, such as lenses, microscopes, and telescopes. Additionally, understanding actual depth and apparent depth can help us to accurately measure the depth of bodies of water and determine the location of objects underwater.

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