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

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

Radius of the curvature is

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

In the object-image relationship input $R=\frac{d}{2}=3,n_2=1.6,n_1=1\mathrm{and}s=∞$

We get

$\frac{1}{∞}+\frac{1.6}{s\text{'}}=\frac{1.6-1}{3}$

And solve for s’

$s\text{'}=8.0cm$

Hence, Image is formed at a distance of 8cm to the right

In the object-image relationship input$R=\frac{d}{2}=3,n_2=1.6,n_1=1ands=12$

We get

$\frac{1}{12}+\frac{1.6}{s\text{'}}=\frac{1.6-1}{3}$

And solve for s’

$s\text{'}=13.7cm$

Hence, Image is formed at a distance of 13.7cm to the right

In the object-image relationship input$R=\frac{d}{2}=3,n_2=1.6,n_1=1ands=12$

We get

$\frac{1}{2}+\frac{1.6}{s\text{'}}=\frac{1.6-1}{3}$

And solve for s’

$s\text{'}=-5.33cm$

Hence, Image is formed at a distance of -5.33cm to the left