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A swimmer is diving in a swimming pool vertically down with a velocity of 2 ms^(-I). What will be the velocity as seen by a stationary fish at the bottom of the pool right below the diver? Refractive index of water is 1.33 |
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Answer» `2.66 cms^(-1)` `therefore(h_i)/(h_0)=(n_("water"))/(n_("air"))` `therefore h_i=h_0xx1.33` From equation (1),`h_i=1.33 h_0` difference with respect to time, `therefore(dh_i)/(dt)=1.33(dh_0)/(dt)` `therefore v_i=1.33 v_0` `=1.33xx2` `=2.66 m//s` So, the fish will see the swimmer falling with speed of 2.66 m`//`s. |
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