A small bulb is placed at the bottom of a tank containing water to a depth of 80cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.)
Given: Actual depth of bulb d1=0.8m The refractive index of water is μ=1.33 Circle's radius, R=2AC We know that, μ=sinisin90∘ ⇒i=48.75∘ Now, in the ΔOBC, tan i=OBOC=d1R⇒R=0.91m Area, πR2=π×(0.91)2=2.61m2
