A small bulb is placed at the bottom of a tank containing water to adepth of 80cm. What is the area of the surface of water throughwhich light from the bulb can emerge out? Refractive index of wateris 1.33. (Consider the bulb to be a point source.)
Given: The depth of the small bulb at the bottom of a tank containing water is 80 cm and the refractive index of water is 1.33 .
Consider the figure shown below.
The refractive index of glass with respect to air is given by Snell’s law,
μ = sin r sin i
Where, the angle of incidence is i and the angle of refraction is r .
By substituting the given values in the above expression, we get
1.33 = sin 90 ° sin i i = sin − 1 ( 1 1.33 ) = 48.75 °
From the figure, in triangle BOC ,
tan i = OC OB
By substituting the given values in the above expression, we get
tan 48.75 ° = OC 0.8 OC = 0.91 m
The bulb is consider as a point source so, the emergent light consider as a circle of radius OC .
The area of the surface of water ( A ) is given as,
A = π ( OC ) 2 = π × ( 0.91 ) 2 = 2.60 m 2
Thus, the area of the surface of water through which light from the bulb can emerge out is 2.60 m 2 .
Given: The depth of the small bulb at the bottom of a tank containing water is
Consider the figure shown below.
The refractive index of glass with respect to air is given by Snell’s law,
Where, the angle of incidence is
By substituting the given values in the above expression, we get
From the figure, in triangle
By substituting the given values in the above expression, we get
The bulb is consider as a point source so, the emergent light consider as a circle of radius
The area of the surface of water
Thus, the area of the surface of water through which light from the bulb can emerge out is
