InterviewSolution
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InterviewSolution
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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. |
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Answer» Solution :`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` `thereforex_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)` |
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