A pole of length `2.00 m` stands half dipped in a swimming pool with level `1 m` higher than bed (bottom). The refractive index of water is `4//3` and sunlight is coming at an angle of `37^(@)` with the vertical. Find the length of the shadow of the pole on the bed. Use `sin^(-1)(0.45)=26.8^(@), tan(26.8^(@))=0.5`.

Answer

`1.25 m`

Explanation

Let in the adjoining figure `AB` be the pole. So, `AD` represents its shadow on the bed. In `DeltaBCE`
`tan 37^(@) =(CE)/(BE) =(y)/(1)`
`rArr y=(3)/(4) m`.
Also due to refraction of sunray at the water surface
`sin 37^(@) =(4)/(3) sinr`
`rArr sin r =(9)/(20)`
`therefore tan r approx 0.5`
So, In `Delta CDF`
we have `tanr =(DF)/(CF)`
`rArr 0.5 =(x)/(1)`
`therefore x=0.5`
`therefore` Length of the shadow `AD=x + y =1.25 m`.