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Setup help needed:apparent depth vs. actual depth 
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#1
Apr2507, 04:41 PM

P: 16

there's a light source, S, sitting 2.60 metres below the surface of a pool 1.00 metres away from one side. i'm supposed to find the angle at which the light left the water and the difference between the apparent and actual depths of the light source.
i attempted using Snell's Law with n of water = 1.33 and the incedent angle = 90 degrees. 1.33(sin90)=1.00(sin_theta) I got theta=1.33. that can't be right. I also tried trig. Theta is the angle of a rt triangle with the light source. 2.60 is the opposite, 1.00 is adjacent. Using the cot_theta, my answer was 69 degrees. neither of these answers was accepted by WebAssign. HELP!!! 


#2
Apr2507, 04:48 PM

P: 16

Also, there is a question as to how to find the critical angle of a "special glass" (the index of refraction of which I am not given... o.O) when submerged in water. Hereis the question:
"The critical angle for a special glass in air is 43°. What is the critical angle if the glass is immersed in water?" I tried to use sin_theta=1/n, but I couldn't figure out how to set up the problem. Someone have an idea? 


#3
Apr2507, 05:23 PM

Sci Advisor
HW Helper
P: 8,954

It depends on where you are looking from.
If you are directly above the light then, as you said i=90deg and the light goes straight out  the apparent depth is then just real_depth/n. If you are standing on the side of the pool then just draw a triangle, and the real depth is the hypotenuse. 


#4
Apr2507, 05:27 PM

P: 16

Setup help needed:apparent depth vs. actual depth



#5
Apr2507, 05:29 PM

P: 16




#6
Apr2507, 05:40 PM

P: 16

sorry, that didn't register... thnx. 


#7
Apr2507, 05:53 PM

P: 16

what i did: Pythag. theorem => hyp=2.78m D(apparant)=D(real)/n D(apparant)=2.78/1.33 D(apparant)=2.09 Difference = .69m This answer was deemed incorrect. Say that ang.B=rt ang. Ang.A=observer's eyes to S (the angle that the light meets the water/air boundary to the normal inside the water). Ang.C=S to ang.B tanA=2.6/1 tanA=2.6 tan1A= 68.96 Should i input this angle into Snell's Law? 


#8
Apr2507, 06:20 PM

HW Helper
P: 1,541




#9
Apr2507, 06:25 PM

P: 16




#10
Apr2507, 06:28 PM

HW Helper
P: 1,541

Well yes, for the special case of a critical angle.



#11
Apr2507, 06:31 PM

P: 16

Tried pitting it in Snell's, hage, but it didn't work... I came out with 38.22 degrees, and used 43 degrees as the incedent angle.
how do i find the crit angle in the water? if i use the glass' n, i'd only get 43 degrees again. i think i need the index of refraction at the water/glass boundary. but i am so confused.... 


#12
Apr2507, 06:37 PM

HW Helper
P: 1,541

You know that n=1.466 for the glass, and n=1.33 for the water (looking it up).



#13
Apr2507, 06:38 PM

P: 16

How do you set up snell's law for a critical angle?! i thought snell's law didn't work with critical angles...



#14
Apr2507, 06:48 PM

HW Helper
P: 1,541

It's a special case. Let theta1 = critical angle, let theta2 = 90 degrees (for total internal reflection).
http://hyperphysics.phyastr.gsu.edu/hbase/hframe.html 


#15
Apr2507, 07:48 PM

P: 16

theta1 with which n, glass or water?



#16
Apr2507, 07:58 PM

P: 16

GOT IT! Thanks, hage!



#17
Apr2507, 07:59 PM

P: 16

got some insight into the other problem of mine?



#18
Apr2507, 08:13 PM

P: 16




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