A jar of height h is filled with a transparent liquid of refractive index μ (Fig). At the centre of the jar on the bottom surface is a dot. Find the minimum diameter of a disc, such that when placed on the top surface symmetrically about the centre, the dot is invisible.
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Key concept:
In the figure, ray 1 strikes the surface at an angle less than critical angle c and gets refracted in rarer medium. Ray 2 srtikes the surface at critical angle and grazes the interface. Ray 3 strikes the surface making an angle greater than the crirical angle and gets internally reflected. The locus of points where ray strikes at critical angle is a circlr, called circle of illuminance (C.O.I). All light rays striking inside the circle of illuminance get refracted in the rarer medium. If an observer is in the rarer mecdium, he/she will see light coming out only from within the circle of illuminance. If a circular opaque plate covers the circle of illuminance, no ligth will get refraced in the rarer medium and then the object cannot be seen from the rarer medium. In figure, O is a smll dot at the bottom of the jar. the ray from the dot emerges out of a circular pathc of water surface of diameter AB till the angle of incidence for the rays OA and OB exceeds the critical angle $(i_C).$
Rays of light incident at an angle greater than $i_C$, are totally reflected within water and consequently cannot emerge out of the water surface. As $\sin\text{i}_\text{C}=\frac{1}{\mu}\Rightarrow\ \tan\text{i}_\text{C}=\frac{1}{\sqrt{\mu^2-1}}$ Now, $\frac{\frac{\text{d}}{2}}{\text{h}}=\tan\text{i}_\text{C}$ $\Rightarrow\ \frac{\text{d}}{2}=\text{h}\tan\text{i}_\text{c}=\text{h}\frac{1}{\sqrt{\mu^2-1}}$ $\Rightarrow\ \text{d}=\frac{2\text{h}}{\sqrt{\mu^2-1}}$ Thsi is the required expression of d.
In the figure, ray 1 strikes the surface at an angle less than critical angle c and gets refracted in rarer medium. Ray 2 srtikes the surface at critical angle and grazes the interface. Ray 3 strikes the surface making an angle greater than the crirical angle and gets internally reflected. The locus of points where ray strikes at critical angle is a circlr, called circle of illuminance (C.O.I). All light rays striking inside the circle of illuminance get refracted in the rarer medium. If an observer is in the rarer mecdium, he/she will see light coming out only from within the circle of illuminance. If a circular opaque plate covers the circle of illuminance, no ligth will get refraced in the rarer medium and then the object cannot be seen from the rarer medium. In figure, O is a smll dot at the bottom of the jar. the ray from the dot emerges out of a circular pathc of water surface of diameter AB till the angle of incidence for the rays OA and OB exceeds the critical angle $(i_C).$
Rays of light incident at an angle greater than $i_C$, are totally reflected within water and consequently cannot emerge out of the water surface. As $\sin\text{i}_\text{C}=\frac{1}{\mu}\Rightarrow\ \tan\text{i}_\text{C}=\frac{1}{\sqrt{\mu^2-1}}$ Now, $\frac{\frac{\text{d}}{2}}{\text{h}}=\tan\text{i}_\text{C}$ $\Rightarrow\ \frac{\text{d}}{2}=\text{h}\tan\text{i}_\text{c}=\text{h}\frac{1}{\sqrt{\mu^2-1}}$ $\Rightarrow\ \text{d}=\frac{2\text{h}}{\sqrt{\mu^2-1}}$ Thsi is the required expression of d.
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