Question

A beaker contains water up to a height $h_1$ and K oil above water up to another height $h_2.$ Find the apparent shift in the position of the bottom of the beaker when viewed from above. Refractive index of water is ${\mu }_1$ and that of K.oil is ${\mu }_2.$

• A. $(1-\frac{1}{{\mu }_1})h_1+(1-\frac{1}{{\mu }_2})h_2.$
• B. $(1-\frac{1}{{\mu }_1})h_1-(1-\frac{1}{{\mu }_2})h_2.$
• C. $(1-\frac{1}{{\mu }_1})h_1+(1-\frac{1}{{\mu }_2})h_2.$
• D. $(1-\frac{1}{{\mu }_1})h_1-(1-\frac{1}{{\mu }_2})h_2.$

Answer 1

The correct option is A

$(1-\frac{1}{{\mu }_1})h_1+(1-\frac{1}{{\mu }_2})h_2.$

The apparent shift of the bottom due to the water is
Refractive index$={\mu }_1,{\mu }_2$
Thickness of the slab$=h$
Vertical shift$=△h$
$\mu =\frac{h}{h-△h}$

$(h-△h)\times \mu =h$

$(\mu -1)\times h=\mu △h$

$△h=(1-\frac{1}{\mu })\times h$
Hence
$△x_1=(1-\frac{1}{{\mu }_1})h_1$
and due to the K.oil is
$△x_2=(1-\frac{1}{{\mu }_2})h_2.$

The total shift = $△x_1+△x_2$
$=(1-\frac{1}{{\mu }_1})h_1+(1-\frac{1}{{\mu }_2})h_2.$