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A Small Piece of Wood is Floating on the Surface of a 2.5 M Deep Lake

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प्रश्न

A small piece of wood is floating on the surface of a 2.5 m deep lake. Where does the shadow form on the bottom when the sum is just setting? Refractive index of water = 4/3.

योग
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उत्तर

Given,
Depth of the lake = 2.5 m
Refractive index (μ) of water = \[\frac{4}{3}\]
When the sun is just setting, θ is approximately = 90˚

Therefore, incidence angle is 90˚
Using Snell's law:
\[\therefore   \frac{\sin  i}{\sin  r} = \frac{\mu_2}{\mu_1}\]
⇒ `1/ sin r = ( 4/3 )/1`
⇒ `r = 49^circ `
From the figure, W'O = x is the distance of the shadow
Thus, `(x/2.5) = tan r = 1.15`
⇒ x=2.5×1.15=2.8 m

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अध्याय 18: Geometrical Optics - Exercise [पृष्ठ ४१३]

APPEARS IN

एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
अध्याय 18 Geometrical Optics
Exercise | Q 16 | पृष्ठ ४१३

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