Three immiscible liquids of densities `d_1 gt d_2 gt d_3` and refractive indies `mu_1 gt mu_2 gt mu_3` are put in a beaker. The light of each liquid column is `(h)/(3)`. A dot is made at the bottom of the beaker. For near normal vision, find the apparent depth of the dot.

Explanation

`rArr` According to formula,
`"Virtual depth " = ("Real depth")/("Refractive index of denser medium w.r.t rarer medium")` .....(1)
`rArr` From equation (i)
(i) For first medium :
`x_(1) = (h_1)/(((mu_1)/(mu_2)))=mu_2/(mu_1)(h/3) (because h_1 = h/3)`.........(2)
(ii) For second medium :
`x_2 = (h_2)/(((mu_2)/(mu_3)))`
`rArr` From figure `h_2 = (h)/(3) + x_1`
`therefore x_2 = (h/3 + x_1)/(mu_1/m_3)= (mu_3)/(mu_2){h/3 + (mu_2/mu_1 xx h/3)}.......(3)`
[ From equation (2)]
(iii) For the third medium
`x_3 = (h_3)/((mu_3/1))`
`rArr` Thus, from figure `h_3 = h/3 xx x_2`
`x_3 = ((h/3 +x_2))/(mu_3)`
`= (1)/(mu_3)[h/2 + mu_3/mu_2{h/3 + (mu_2/mu_1 xx h/3)}]`
[From equation (3)]
`= 1/(mu_3)(h/3) +(1)/(mu_2){h/3 + (mu_2/mu_1 xx h/3)}`
`= 1/mu_3(h/3) + 1/mu_2(h/3) + 1/(mu_1)(h/3)`
`therefore x_3 = h/3(1/mu_1 + 1/mu_2+1/m_3)`