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InterviewSolution
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A bulb is placed at a depth of 2sqrt(7) m in water and a floating opaque disc is placed over the bulb so that the bulb is not visible from the surface. What is the minimum diameter of the disc? |
Answer» Solution : As shown in figure, LIGHT from bulb will not emerge out of water if at the edge of disc, `i gt theta_cor sintheta_c` Now if R is the radius of disc and h is the depth of bulb from it, `sin i = (R )/(sqrt( R^2 +h^2)) ` and` sin theta_c = (1)/(mu )` So equation (1) becomes `(R )/( sqrt(R^2+h^2))gt(1)/(mu)orR gt(h )/( sqrt(mu^2-1)) ` here `h=2 sqrt(7)mand mu= ( 4//3)` So `R_(min) = ( 2 sqrt(7))/( sqrt((16 //9)-1))= 6M ` So DIAMETER of disc = 2R = 2 `xx` 6 = 12 m |
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