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Question
Solve Numerical example.
A monochromatic ray of light strike the water (n = 4/3) surface in a cylindrical vessel at angle of incidence 53°. Depth of water is 36 cm. After striking the water surface, how long will the light take to reach the bottom of the vessel? [Angles of the most popular Pythagorean triangle of sides in the ratio 3:4:5 are nearly 37°, 53°, and 90°]
Solution
From figure,
ray PO = incident ray
ray OA = refracted ray
QOB = Normal to the water surface.
Given that, ∠POQ = angle of incidence (θ1) = 53°
Seg OB = 36 cm and nwater = `4/3`
From Snell’s law,
n1 sinθ1 = n2 sinθ2
∴ nwater = `sinθ_1/sinθ_2`
Or sinθ2 = `sinθ_1/"n"_"water"=(sin(53°)xx3)/4`
∴ θ2 ≈ 37°
ΔOBA forms a Pythagorean triangle with angles 53°, 37°, and 90°.
Thus, sides of ΔOBA will be in ratio 3: 4: 5
Such that OA forms the hypotenuse. From the figure, we can infer that,
Seg OB = 4x = 36 cm
∴ x = 9
∴ seg OA = 5x = 45 cm and
seg AB = 3x = 27 cm.
This means the light has to travel 45 cm to reach the bottom of the vessel.
The speed of the light in water is given by,
v = `"c"/"n"`
∴ v = `(3xx10^8)/(4//3)=9/4xx10^8`m/s
∴ Time taken by light to reach the bottom of vessel is,
t = `"s"/"v"=(45xx10^-2)/(9/4xx10^8)` = 20 × 10−10 = 2 ns or 0.002 μs
The light will take 2 ns or 0.002 μs to reach the bottom of the vessel
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