System Configuration

We have a steel wire of diameter $d$ embedded coaxially (along the axis) of a glass tube. We view this wire from outside the tube.

• Refractive index of glass (medium 1): $n_1 = 4/3$
• Refractive index of air (medium 2): $n_2 = 1$
• Apparent diameter of the wire: $d’ = 8/3 \text{ mm}$

Let the radius of the glass tube be $R$. Since the wire is coaxial, the distance of the object (wire) from the refracting surface is equal to the radius of the tube.

Refraction at a Cylindrical Surface

We apply the refraction formula for a curved surface:

$$ \frac{n_2}{v} – \frac{n_1}{u} = \frac{n_2 – n_1}{R_s} $$

Using the sign convention (light travels from object to observer):

• $u = -R$ (Object is at the center, light travels outward to surface at distance R)
• $R_s = -R$ (Center of curvature is inside, towards the negative side of the pole)

Substituting the values:

$$ \frac{1}{v} – \frac{4/3}{-R} = \frac{1 – 4/3}{-R} $$ $$ \frac{1}{v} + \frac{4}{3R} = \frac{-1/3}{-R} $$ $$ \frac{1}{v} + \frac{4}{3R} = \frac{1}{3R} $$ $$ \frac{1}{v} = \frac{1}{3R} – \frac{4}{3R} = -\frac{3}{3R} = -\frac{1}{R} $$ $$ v = -R $$

This result implies that the image is formed at the same location as the object (at the center of the cylinder).

Lateral Magnification

For a single refracting surface, the lateral magnification $m$ is given by:

$$ m = \frac{h_i}{h_o} = \frac{n_1 v}{n_2 u} $$

Substitute $v = -R$, $u = -R$, $n_1 = 4/3$, and $n_2 = 1$:

$$ m = \frac{(4/3)(-R)}{(1)(-R)} = \frac{4}{3} $$

The magnification is $4/3$. This relates the actual diameter $d$ to the apparent diameter $d’$:

$$ d’ = m \cdot d $$ $$ \frac{8}{3} \text{ mm} = \frac{4}{3} d $$ $$ d = \frac{8/3}{4/3} = 2 \text{ mm} $$