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.)

Solution

Here, the source of light (S) is 80 cm below the surface of water.

i.e., SO = 80 cm = 0.8 m

When angle of incidence,

$∠ i = C, for SA and SB and ∠ r = 90 °$

Therefore, Area of the surface of water through which light from the bulb can emerge is area of the circle of radius,

$r = \frac{AB}{2} = OA = OB As, μ = \frac{1}{sinc} sinc = \frac{1}{μ} = \frac{1}{1.33} = 0.75 C = 48.6 ° In ∆ OBS, \tan C = \frac{OB}{OS} ∴ OB = OS tanc = 0.8 \tan 48.6 ° r = 0.8 \times 1.1345 = 0.907 m So, area of the surface water = {πr}^{2} = 3.14 (0.907)^{2} = 25.18 m^{2}$

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Wave Optics

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.)

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