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Three immiscible liquids of densities d_{1}>d_{2}>d_{3}  and refractive indices \mu _{1}>\mu_{2}>\mu_{3}  are put in a beaker. The height of each liquid column is \frac{h}{3}. A dot is made at the bottom of the beaker. For near normal vision, find the apparent depth of the dot.

 

Answers (1)

Key concept : Coordinate convention: At the first surface (+upward and -ve downward)

\frac{\mu _{2}}{h'}-\frac{\mu _{1}}{(-h)}=\frac{\mu _{2}-\mu _{1}}{\infty } (infinity because the surface is plane)

or h'=\frac{\mu _{2}}{\mu _{1}}h. The negative sign shows that it is on the side of the object

h' is the apparent depth of O after refraction from interface.

The position of image O after the refraction from surface-1. If seen from u_{2}, the apparent depth is h_{1}

h_{1}=\frac{\mu _{2}}{\mu _{1}}\frac{h}{3}

The negative sign indicates that it is on the side of the object.

Since the image formed by surface-1. It will act as an object for surface-2. If seen from u_{3} , the apparent depth is h_{2} .

Similarly, the image formed by medium 2, O_{2} acts as an object for medium 3

h_{2}=-\frac{\mu _{3}}{\mu _{2}}\left (\frac{\mu _{2}}{\mu _{1}}\frac{h}{3}+\frac{h}{3} \right )=-\frac{h}{3}\left (\frac{\mu _{3}}{\mu _{2}}+\frac{\mu _{2}}{\mu _{1}} \right )

Finally the image formed by surface -2 will act as an object for surface-2. If seen from outside, the apparent depth is h_{3}

h_{3}=-\frac{1}{\mu _{3}}\left [\frac{h}{3}+ \frac{h}{3}\left (\frac{\mu _{3}}{\mu _{2}}+\frac{\mu _{2}}{\mu _{1}} \right ) \right ]=-\frac{h}{3}\left ( \frac{1}{\mu _{1}}+\frac{1}{\mu _{2}}+\frac{1}{\mu _{3}} \right )

Hence apparent depth of odt is \frac{h}{3}\left ( \frac{1}{\mu _{1}}+\frac{1}{\mu _{2}}+\frac{1}{\mu _{3}} \right )

Important point:

Apparent depth (distance of final image from final surface)

                  =\frac{t_{1}}{n_{1\; real}}+\frac{t_{2}}{n_{2\; real}}+\frac{t_{3}}{n_{3\; real}}+...+\frac{t_{n}}{n_{n\; real}}

                 

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