CUET Preparation Today
A glass beaker contains water upto a height $h_1$ and kerosene oil above water upto height $h_2$. If the refractive index of water is $\mu_{w}$ and that of oil is $\mu_{k}$, then what will be the apparent shift in the position of the bottom of the beaker when viewed from above? |
$\left[\left(1-\frac{1}{\mu_{w}}\right)+\left(1-\frac{1}{\mu_{k}}\right)\right]\left(h_1+h_2\right)$ $\left(1-\frac{1}{\mu_{k}}\right) h_1+\left(1-\frac{1}{\mu_{w}}\right) h_1$ $\left(1-\frac{1}{\mu_w}\right) h_2+\left(1-\frac{1}{\mu_k}\right) h_1$ $\left(1-\frac{1}{\mu_{w}}\right) h_1+\left(1-\frac{1}{\mu_{k}}\right) h_2$ |
$\left(1-\frac{1}{\mu_{w}}\right) h_1+\left(1-\frac{1}{\mu_{k}}\right) h_2$ |
The correct answer is Option (4) → $\left(1-\frac{1}{\mu_{w}}\right) h_1+\left(1-\frac{1}{\mu_{k}}\right) h_2$ The apparent shift (Δh) due to a layer of thickness h and the refractive index (μ) is - $Δh=h\left(1-\frac{1}{μ}\right)$ $∴Δh_{total}=Δh_{water}+Δh_{oil}$ $=h_1\left(1-\frac{1}{μ_w}\right)+h_2\left(1-\frac{1}{μ_k}\right)$ |
