Q26E - University Physics with Modern Physics

Question

The glass rod of Exercise 34.25 is immersed in a liquid. An object 14.0 cm from the vertex of the left end of the rod and on its axis is imaged at a point 9.00 cm from the vertex inside the liquid. What is the index of refraction of the liquid?

Step-by-Step Solution

Answer

The refractive index of the liquid is 1.244

Step 1: Image-Object relation

Object-Image relationship for spherical reflecting surfaces:

$\frac{n_1}{s}+\frac{n_2}{s}=\frac{n_2-n_1}{R}$

 

$n_1,n_2=>$The refraction index of both the surfaces,$s^{\text{'}}=>$ the image distance from the vertex of the spherical surfaces,  $s=>$ the object distance from the vertex of the spherical surface, $R=>$ and the radius of the spherical surface.

$$\begin{gathered} n_1=required=>liquid \\ {n}_2=1.6\left(glass\right) \end{gathered}$$

s = 14cm

$s^{\text{'}}=-9cm$

R is negative because it is in the opposite direction of refracted light

R = - 4cm

Step 2: Calculation

$$\begin{gathered} \frac{n_1}{s}+\frac{n_1}{s\text{'}}=\frac{n_2-n_1}{R}=\frac{n_2}{R}\frac{n_1}{R} \\ \frac{n_1}{s}+\frac{n_1}{R}=\frac{n_2}{R}-{\frac{n_2}{s\text{'}}}^{\text{'}} \\ {n}_1\left(\frac{1}{s}+\frac{1}{R}\right)=\frac{n_2}{R}-\frac{n_2}{s\text{'}} \\ {n}_1=\left(\frac{n_2}{s}-\frac{n_2}{s\text{'}}\right)/\left(\frac{1}{s}+\frac{1}{R}\right) \\ =\left(\frac{1.6}{-4cm}-\frac{1.6}{-9cm}\right)/\left(\frac{1}{14cm}+\frac{1}{-4cm}\right)=1.244 \end{gathered}$$

Hence, the refractive index of the liquid is 1.244