Solar rays are incident at 450 on the surface of water (μ=4/3) The length of the shadow of a pole of length 1.2 m formed at the bottom of the pond is 3.3n where ‘n’ is. (If the pole is vertical assuming that 0.2 m of the pole is above the water surface)

μ=sin⁡45sin⁡r⇒43=12sin⁡r⇒sin⁡r=342 From this, tan⁡r=323=x1 ⇒x=323≃0.625shadow of pole at the bottom of tank is 0.2+x≃0.825=3.3n⇒n=4

Solar rays are incident at 450 on the surface of water (μ=4/3) The length of the shadow of a pole of length 1.2 m formed at the bottom of the pond is 3.3n where ‘n’ is. (If the pole is vertical assuming that 0.2 m of the pole is above the water surface)

μ=sin⁡45sin⁡r⇒43=12sin⁡r⇒sin⁡r=342 From this, tan⁡r=323=x1 ⇒x=323≃0.625shadow of pole at the bottom of tank is 0.2+x≃0.825=3.3n⇒n=4

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Solar rays are incident at 45⁰ on the surface of water $(μ = 4 / 3)$ The length of the shadow of a pole of length 1.2 m formed at the bottom of the pond is $\frac{3.3}{n}$ where ‘n’ is. (If the pole is vertical assuming that 0.2 m of the pole is above the water surface)

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Detailed Solution

$μ = \frac{\sin 45}{\sin r} \Rightarrow \frac{4}{3} = \frac{1}{\sqrt{2} \sin r} \Rightarrow \sin r = \frac{3}{4 \sqrt{2}} From this, \tan r = \frac{3}{\sqrt{23}} = \frac{x}{1} \Rightarrow x = \frac{3}{\sqrt{23}} ≃ 0.625$
shadow of pole at the bottom of tank is
$0 .2 + x ≃ 0.825 = \frac{3 .3}{n} \Rightarrow n = 4$