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Home » 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.)
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.)
CBSE | Class 12 | NCERT | Physics | Ray Optics and Optical Instruments
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.)

Answer :

We are given the following information –

Bulb’s actual depth in water is d1=80 cm =0.8 m

Refractive Index of water is \mu= 1.33

I is the Angle of incidence

r is the Angle of refraction – 90°

The emergent light is deemed a circle because the bulb is employed as a point source.

R=\frac{AC}{2}=AO=OB

According to Snell’s law, we can express the relation for the refractive index of water in the following manner –

\mu=\frac{sin\;r}{sin\;i}

1.33=\frac{sin\;90°}{sin\;i}

Therefore\;i=sin^{-1}(\frac{1}{1.33})=48.75°

The relation is –

tan i=\frac{OC}{OB}=\frac{R}{d_{1}}

Therefore R=tan\;48.75° \times 0.8=0.91m

Area of the surface of water becomes –

=nR^{2}=n(0.91)^{2}=2.61m^{2}

 2.61m^{2} is the area of water through which the light from the water can emerge out.

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