Question

the bottom of a beaker containing a liquid appears to rise by 4 cm. on increasing the depth of a liquid by 12 cm,the bottom appears to rise by 7 cm. the refractive index of the liquid is??​

Answer 1

Given:

The bottom of a beaker containing a liquid appears to rise by 4 cm. On increasing the depth of a liquid by 12 cm, the bottom appears to rise by 7 cm.

To find:

Refractive Index of liquid.

Calculation:

$\sf{rise \: in \: image = real \: depth - apparent \: depth}$

$\sf{ = > r= d - \dfrac{d}{ \mu} }$

$\sf{ = > r= d (1- \dfrac{1}{ \mu} )}$

In 1st case:

$\sf{ 4= d (1- \dfrac{1}{ \mu} ) \: \: \: \: \: ........(1)}$

In 2nd case:

$\sf{ 7= (d + 12)(1- \dfrac{1}{ \mu} ) \: \: \: \: \: ........(2)}$

Dividing the two Equations:

$\rm{ \therefore \dfrac{d + 12}{d} = \dfrac{7}{4} }$

$\rm{ = > 4d + 48 = 7d}$

$\rm{ = >3d = 48 }$

$\rm{ = >d = 16 \: m }$

Now, putting value of d in eq.(1):

$\sf{ \therefore 4= 16 \times (1- \dfrac{1}{ \mu} ) }$

$\sf{ = > \dfrac{1}{4} = (1- \dfrac{1}{ \mu} ) }$

$\sf{ = > \dfrac{1}{ \mu} = 1 - \dfrac{1}{4} }$

$\sf{ = > \dfrac{1}{ \mu} = \dfrac{3}{4} }$

$\sf{ = > \mu= \dfrac{4}{3} }$

So, final answer is:

$\boxed{ \mathfrak{ \red{ \mu= \dfrac{4}{3} }}}$

Answer 2

4/3 answer

hope it helful