Q23E - University Physics with Modern Physics

Question

The glass rod of Exercise 34.22 is immersed in oil (n = 1.452). An object placed to the left of the rod on the rod’s axis is to be imaged 1.20 m inside the rod. How far from the left end of the rod must the object be located to form the image?

Step-by-Step Solution

Answer

Object must be 39.5cm from the left end of the rod

Step 1: Important Concepts

The object image relationship for a spherical reflecting surface is 

$\frac{{\mathbf{n}}_{\mathbf{1}}}{\mathbf{s}}\mathbf{+}\frac{{\mathbf{n}}_{\mathbf{2}}}{\mathbf{s}\mathbf{\text{'}}}\mathbf{=}\frac{{\mathbf{n}}_{\mathbf{2}}\mathbf{-}{\mathbf{n}}_{\mathbf{1}}}{\mathbf{R}}$     

Where ${\mathit{n}}_{\mathbf{2}}$ and ${\mathit{n}}_{\mathbf{1}}$ are the refraction index of the surfaces 

s is th object distance from the vertex of the spherical surface and 

$\mathit{s}\mathbf{\text{'}}$ is the image distance from the vertex of the spherical surface and  is the radius of curvature of the spherical surface

Magnification is given by 

$\mathit{m}\mathbf{=}\mathbf{-}\frac{{\mathbf{n}}_{\mathbf{1}}\mathbf{s}\mathbf{\text{'}}}{{\mathbf{n}}_{\mathbf{2}}\mathbf{s}}$                                                        

 

Radius of the curvature is 

$\mathit{R}\mathbf{=}\frac{\mathbf{d}}{\mathbf{2}}$                                                           

Step 2: Find the object distance

Here we have 

$$\begin{gathered} n_2=1.6\to For oil \\ {n}_1=1.45 \end{gathered}$$                                                                                                               

                                                           

The image is located at 

$s\text{'}=1.2m$                                                           

It is negative since the image is on the opposite side of the outgoing light.

 

We know since the surface of water-air interaction is planar hence $R=\infty$

 

Input this information in the formula 

 

$$\begin{gathered} \frac{n_1}{s}+\frac{n_2}{s\text{'}}=\frac{n_2-n_1}{0.03} \\ \frac{1.45}{s}+\frac{1.6}{1.2}=\frac{1.6-1.45}{0.03} \end{gathered}$$                                                           

 

 

Rearranging we have 

$$\begin{gathered} s=n_1\times \frac{3}{11} \\ s=1.45\times \frac{3}{11} \\ s=0.395m=39.5cm \end{gathered}$$                                                     

 

 

 

Hence,Object must be 39.5cm from the left end of the rod