Q24E - University Physics with Modern Physics

Question

The left end of a long glass rod 8.00 cm in diameter, with an index of refraction of 1.60, is ground and polished to a convex hemispherical surface with a radius of 4.00 cm. An object in the form of an arrow 1.50 mm tall, at right angles to the axis of the rod, is located on the axis 24.0 cm to the left of the vertex of the convex surface. Find the position and height of the image of the
arrow formed by paraxial rays incident on the convex surface. Is the image erect or inverted?

Step-by-Step Solution

Answer

A. Image is formed at the a distance of 14.77cm

B. Height of the image is 0.577mm and it is inverted

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 R is the radius of curvature of the spherical surface

Magnification is given by 

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

Step 2: Find the object distance

Here we have                                                                                                                       

                                         $$\begin{gathered} n_2=1.6\to For \mathrm{glass} \\ {n}_1=1 \end{gathered}$$  

The image is located at 

                                                            

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

$R=0.04m$ 

Input this information in the formula 

 

 $$\begin{gathered} \frac{n_1}{s}+\frac{n_2}{s\text{'}}=\frac{n_2-n_1}{R} \\ \frac{1}{0.24}+\frac{1.6}{s\text{'}}=\frac{1.6-1}{0.04} \end{gathered}$$                                                           

 

 

Rearranging we have

$$\begin{gathered} s\text{'}=\frac{1.6}{{\displaystyle \frac{1.6-1}{0.04}}-{\displaystyle \frac{1}{0.04}}} \\ s\text{'}=0.1477m \\ s\text{'}=14.77cm \end{gathered}$$                                                     

Hence, Image is formed at the a distance of 14.77cm

Step 3: Find the magnification

Linear magnification is given by 

$m=-\frac{n_{1}s\text{'}}{n_{2}s}=\frac{y\text{'}}{y}$                                               

                                                    

Where $y\text{'}$ is the height of the image, and y is the height of the object

 

Rearrange to get 

                                                $$\begin{gathered} y\text{'}=-\frac{n_{1}s\text{'}y}{n_{2}s} \\ y\text{'}=-\frac{\left(1\right)\left(0.1477m\right)\left(0.0015\right)}{1.6\left(0.24m\right)}=-5.77mm \end{gathered}$$

 

Hence,Height of the image is 0.577mm and it is inverted