If the refractive index of water is 4/3 and th fish is 12cm below the surface of water, find the radius of the circle.

Question

A fish looking up through the water sees that the outside world is contained in a circular horizon. If the refractive index of water is 4/3 and th fish is 12cm below the surface of water, find the radius of the circle.

Answer 1

sini ( in water)=3/4 = 12/(r² +12² )^1/2

If r is the radius of the circle in which fish can view

 9/16 = 144/(r² +144)

r² +144 =144x16/9

 

r² = (144x16/9) -144

 

r² = 144(16/9 -1)

    =144 (16-9)/9

    =144x7/9

    =112

 r = (112) ^1/2

   = 10.58cm

Answer 2

Hi Ashish!

 

What this questions really boils down to is "what is the critical angle for a ray of light going from water to air?"  After that, a little geometry will get us to the radius of the circle.

 

The critical angle is the angle at which light, coming from the water, will have a 90º refraction angle going into the air; i.e. it will travel exactly along the surface boundary between the air and water.  If light coming from the water encounters the boundary at an incident angle greater than the critical angle, it will "totally internally reflect" back into the water, and there will be no (real) refraction.

 

To solve for the critical angle, θ_c, use Snell's law with a refraction angle of 90º.  In this equation, Medium 1 will be the "incident" medium, where the ray comes from (in this case, the water), and Medium 2 will be the "refracted" medium, where the ray is refracted into (the air, for this case).

 

n₁sin(θ_c) = n₂sin(90^o)

 

[NOTE:  both angles are measured with respect to the normal, a line perpendicular to the boundary between the media]

 

Since sin(90º) = 1, this reduces to:

 

sin(θ_c) = n₂/n₁

 

[NOTE: since sin(θ) can only take on values ≤1, n₁ must be greater than n₂ for a critical angle to exist; thus, a critical angle (and total internal reflection) only occurs when going from a higher-index medium, like the water in this example, to a lower-index medium, like the air here]

 

Solve for θ_c using inverse sine.

 

This means that any ray traveling from the water into the air that encounters the boundary at an angle of θ_c will refract exactly along the boundary.  Imagine turning this around; the situation is exactly the same, just in the opposite order.  A ray coming from the air nearly parallel to the boundary will refract into the water at very nearly the critical angle.  This means the critical angle is the boundary case for light being able to refract from the air into the water.

 

This next part may be tricky to see without a diagram, so I apologize.  If you draw a line from the fish directly upward, and then draw, from the end of that line, the radius of the fish's circular viewing horizon on the surface of the water, then draw a line from the end of that previous line back to the fish, you create a right triangle, with the right angle between the vertical line and the radius on the water.  The angle of that triangle between the vertical line and the hypotenuse is then the maximum angle from the vertical at which light from the air will reach the fish.  This angle, by geometric axiom, is congruent to the refraction angle of light coming from the air into the water.

 

Since the maximum possible angle at which light from the air can refract into the water is θ_c from above, that angle between the vertical and the hypotenuse in the right triangle must also be θ_c.  You now have a right triangle where you know one acute angle and the length of the side adjacent to that angle (12 cm), and you want to solve for the opposite side (the horizon radius).  You can do that with trigonometry.

 

I hope that helps!  I know it can be tricky to see without a drawing, so do not hesitate to ask more questions.  Also, let me know if you would like to check an answer.