Question

The left end of a long glass rod 6.00 cm in diameter has a convex hemispherical surface 3.00 cm in radius. The refractive index of the glass is 1.60. Determine the position of the image if an object is placed in air on the axis of the rod at the following distances to the left of the vertex of the curved end: (a) infinitely far, (b) 12.0 cm; (c) 2.00 cm.

Short answer

A. Image is formed at a distance of 8cm to the right

B.Image is formed at a distance of 13.7cm to the right

C. Image is formed at a distance of -5.33cm to the left

Step-by-step solution

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 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

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

When object is at infinity

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

We get

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

And solve for s’

$s\text{'}=8.0cm$

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

When object is at 12cm

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

When object is at 2cm

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