91Ó°ÊÓ

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 andR is the radius of curvature of the spherical surface

Magnification is given by

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

Here we have

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

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

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