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A vessel is filled with two mutually immiscible liquids of refractive indices mu^(1) and mu^(2). The depths of the two liquids are d_(1) and d_(2) respectively. There is a mark at the bottom of the vessel. Show that the apparent depth of the mark when viewed normally is given by ((d_(1))/(mu_(1))+(d_(2))/(mu_(2))). |
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Answer» Solution :The image of p is formed at Q due of refraction at the surface of SEPARATION B of the 1st and 2ND liquid. Another final image due to refraction in air from the second liquid is formed at R. For the first refraction, `(mu_1)/(mu_(2)) = (BP)/(BQ)` `or, "" BQ = (mu_(2))/(mu_(1)).BP` `(mu_(1))/(mu_2)d_(1)` For the second refraction, `(mu_(2))/(1) = (AQ)/(AR)` `or, "" AR = (AQ)/(AR) = (1)/(mu_(2)) (AB + BQ)` `or, "" AR = (1)/(mu_(2))(d_(2) + (mu_(2))/(mu_(1))d_(1)) = (d_(2))/(mu_(2)) + (d_(1))/(mu_(1))` `therefore` APPARENT depth of the mark P when viewed NORMALLY `= AR = (d_(1))/(mu_(1)) + (d_(2))/(mu_(2))` 
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