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

$\mu t$ the real depth =$=x$

Shift = 4cm

$\text{shift}=\left(1-\frac{1}{\mu }\right)x$

$4=\left(1-\frac{1}{\mu }\right)x\to \left(1\right)$

$\Rightarrow x=\frac{4\mu }{\mu -1}$

Now R.D $=x+12$

The shift = 7

$7=\left(1-\frac{1}{\mu }\right)\left(x=12\right)\to \left(2\right)$

Substitute $x=\frac{4\mu }{\mu -1}\text{in}\&-\left(2\right)$

$7=\left(1-\frac{1}{\mu }\right)\left(\frac{4\mu }{\mu -1}+12\right)$

$7=\left(\frac{\mu -1}{\mu }\right)\left(\frac{16\mu -12}{\mu -1}\right)$

$7=\frac{16\mu -12}{\mu }$

$7\mu =16\mu -12$

$9\mu =12$

$\mu =\frac{12}{9}$

$\mu =\frac{4}{3}$