Refraction

[Function]

Hello

 

Another question of my med school approval exam:

(Really don't know how to begin this one)

 

A fish swims 80 cm below surface. The breaking index of water is 1.33. The depth at which a person who looks perpendicular on to the water, right above the fish, sees this fish is

A. 33 cm

B. 60 cm

C. 75 cm

D. 107 cm

 

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So I began with Snell's law: [math]n_1\cdot\sin{\theta_1}=n_2\cdot\sin{\theta_2}[/math]

[math]\Leftrightarrow -1=1.33\cdot\sin{\theta_2}[/math]

[math]\Leftrightarrow\sin{\theta_2}\approx -\frac{3}{4}[/math]

 

So [math]\theta_2 \approx 50^{\circ}[/math]

 

But I don't know if I am something with this?

 

Could someone help me on this one?

 

I might think that I could pick the point (0;-0.8) and rotate this over an angle 90°-arcsin(-3/4) until it hits the refracted 'light ray'?

 

Thanks.

 

F.

Edited April 20, 2014 by Function

[studiot]

You are talking about the phenomenon called 'real and apparent depth'.

 

Look here

 

http://www.physicstutorials.org/home/optics/refraction-of-light/apparent-depth-real-depth

[Enthalpy]

Something is wrong in the 50° because perpendicular to the surface at one side remains perpendicular at the other. Probably the reference of the angle.

 

Instead of learning whether it's a sine or cosine and what sign it has and from where the angle is measured, and so on, it's better to remember simple ideas, like "the speed component parallel to the interface matches on both sides", and deduce the formula each time you need it. You'd make fewer mistakes.

 

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I don't quite understand "the depth at which one sees the fish". If looking vertically, one sees the fish below, and nothing more. Shall the observer have lidar eyes that measure a propagation speed to deduce a distance?