1.

A mark at the bottom of a liquid beaker appears to rise by 0.1 m. If the depth of the liquid is 1.0 m, then refractive index of the liquid is ......

Answer»

1.1
1.3
1.5
1.7

Solution :`("VIRTUAL depth h.")/("REAL depth h")=(n_("AIR"))/(n_("water"))`
(h=1 m, h.=1-0.1=0.9 m,`n_("air")`=1)
`THEREFORE n_w=(hxxn_("air"))/(h.)=(1xx1)/(0.9)=1.1`


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