A fish views outside from a depth h inside water of refractive index n, then the diameter of the c circle through which the outer objects become visi ble will beircle
Answers
Answer:
As the fish sees through the diameter of the circle hence the radius be r and the depth from the top to the fish be h. The fish sees through an angle of C.
So, tanC = r/h.
So, r= h*tanC.
So, r=h*sinC/cosC.
So, r = hsinC/√(1-sin^2C).
Now, we know that the value of sinC will be 1/n, where n is the refractive index of water.
So, r = h*1/n / √(1 - 1/n^2).
So, r = h / √(n^2 - 1).
So, the diameter of the circle will be twice the radius.
Diameter = 2h/√(n^2 - 1).
Answer:
2r/sqrt(n^2 - 1)
Explanation:
We have depth as 'h'
We can take radius to be 'r'
Then, we have TIR = i_c
So tani_c = Perpendicular/Base
Here, we can se that 'r' will become the perpendicular and 'h' will be the base
So, tan i_c = r/h
r = h*tan i_c
[We know the property, tan i_c = sin i_c/ cos i_c]
r = h*sin i_c/cos i_c
[We know the property sin^2 x + cos^2 x = 1 => cos^2 x = 1 - sin^2 x
=> cos x = sqrt(1 - sin^2 x)]
r = h* sin i_c/ sqrt(1-in^2 x)
[We know the property, sin i_c = 1/n]
r = (h*1/n)/sqrt(1-1/n^2)
r = (h/n)/sqrt{(n^2-1)/n^2}
r = (h/n)/(sqrt)(n^2 - 1)/n ---> The 'n' in denominators would cancel
r = h/sqrt(n^2 - 1)
d = 2r = 2(h/sqrt(n^2-1))
d = 2h/sqrt(n^2 - 1)