Question

Looking into a beaker a student observes a corner of the bottom. Filling the beaker with oil lets him observe the center of the bottom. What is the index of refraction n of the oil? Diameter of the beaker and its height are the same.

Answer 1

Diagram n1 n2

Diameter and height

$d=h=L$

the incident angle for.

$tan\Theta _{1}=\frac{h}{d}=\frac{L}{L}=1\Rightarrow \Theta _{1}=tan^{-1}\left ( 1 \right )=45^{0}$angle of incident light

refractive index of air

$n_{1}=1$

angle to the vertical, of the refracted ray.

$tan\Theta _{2}=\frac{\frac{L}{2}}{L}=\frac{1}{2}\Rightarrow \Theta _{2}=tan\left ( \frac{1}{2} \right )=26.6^{0}$

Refractive index unknown.

$n_{2}$

Applying the law of snell

$sin\Theta _{1}\cdot n_{1}=sin\Theta _{2}\cdot n_{2}$

expression for $n_{2}$

$sin\Theta _{1}\cdot n_{1}=sin\Theta _{2}\cdot n_{2}\Rightarrow n_{2}=n _{1}\cdot \frac{ sin\Theta _{1}}{ sin\Theta _{2}}$

evaluated numerically

$n_{2}=1\cdot \frac{ sin\left ( 45^{0} \right )}{ sin\left ( 26.6^{0} \right )}=1.58$

Finally.

$n_{2}=1.58$