Solution to Question 1: Refraction through a Cylindrical Rod
(a) Explanation of the Appearance
The cylindrical glass rod acts as a cylindrical lens. This type of lens has a focusing (and magnifying) power only in the direction perpendicular to its axis. There is no curvature—and therefore no refraction or magnification—along the direction parallel to its axis.
Consider a ruled line on the paper. We can resolve the length vector of any segment of this line into two components:
- Parallel Component ($x_{\parallel}$): Parallel to the rod’s axis. This component remains unchanged ($m_{\parallel} = 1$).
- Perpendicular Component ($x_{\perp}$): Perpendicular to the rod’s axis. Due to the cylindrical curvature, this component is magnified by a factor $m$ ($x’_{\perp} = m \cdot x_{\perp}$).
Because one component is magnified while the other remains constant, the resultant vector (the image of the line) points in a new direction. This causes the lines to appear broken and tilted by an angle $\alpha$ relative to their original direction.
(b) Finding the Refractive Index ($\mu$)
Step 1: Calculate Transverse Magnification ($m$)
Consider the rod as a thick lens viewed from the top. The “object” (the ruled line) is in contact with the bottom surface of the rod.
Let the radius of the rod be $R$.
- Object distance ($u$): The line is at the bottom of the diameter, so $u = -2R$ (sign convention: light travels up).
- Refraction at top surface: Light travels from glass ($\mu_1 = \mu$) to air ($\mu_2 = 1$). The radius of curvature for the top surface is $R$ (center of curvature is below, $C = -R$).
Using the single surface refraction formula $\frac{\mu_2}{v} – \frac{\mu_1}{u} = \frac{\mu_2 – \mu_1}{R_{surf}}$:
$$ \frac{1}{v} – \frac{\mu}{-2R} = \frac{1 – \mu}{-R} $$
$$ \frac{1}{v} = \frac{\mu – 1}{R} – \frac{\mu}{2R} = \frac{2\mu – 2 – \mu}{2R} = \frac{\mu – 2}{2R} $$
$$ v = \frac{2R}{\mu – 2} $$
The magnification $m$ for a single spherical surface is given by $m = \frac{\mu_1 v}{\mu_2 u}$:
$$ m = \frac{\mu \cdot \left(\frac{2R}{\mu – 2}\right)}{1 \cdot (-2R)} = \frac{\mu}{2 – \mu} $$
Step 2: Relate Magnification to Angles
Let the angle of the ruled lines with the rod’s axis be $\theta$. The slope of the line with respect to the axis is $\tan \theta = \frac{x_{\perp}}{y_{\parallel}}$.
The image line is tilted by an additional angle $\alpha$, so its total angle with the axis is $\theta + \alpha$. The new slope is:
$$ \tan(\theta + \alpha) = \frac{x’_{\perp}}{y’_{\parallel}} $$
Since $x’_{\perp} = m x_{\perp}$ and $y’_{\parallel} = y_{\parallel}$, we have:
$$ \tan(\theta + \alpha) = \frac{m x_{\perp}}{y_{\parallel}} = m \tan \theta $$
$$ m = \frac{\tan(\theta + \alpha)}{\tan \theta} $$
Step 3: Solve for $\mu$
Equating the two expressions for $m$:
$$ \frac{\mu}{2 – \mu} = \frac{\tan(\theta + \alpha)}{\tan \theta} $$
Rearranging to solve for $\mu$:
$$ \mu \tan \theta = (2 – \mu) \tan(\theta + \alpha) $$
$$ \mu \tan \theta = 2 \tan(\theta + \alpha) – \mu \tan(\theta + \alpha) $$
$$ \mu [\tan \theta + \tan(\theta + \alpha)] = 2 \tan(\theta + \alpha) $$
$$ \mu = \frac{2 \tan(\theta + \alpha)}{\tan \theta + \tan(\theta + \alpha)} $$
(a) Explanation of the Appearance
The cylindrical glass rod acts as a cylindrical lens. This type of lens has a focusing (and magnifying) power only in the direction perpendicular to its axis. There is no curvature—and therefore no refraction or magnification—along the direction parallel to its axis.
Consider a ruled line on the paper. We can resolve the length vector of any segment of this line into two components:
- Parallel Component ($x_{\parallel}$): Parallel to the rod’s axis. This component remains unchanged ($m_{\parallel} = 1$).
- Perpendicular Component ($x_{\perp}$): Perpendicular to the rod’s axis. Due to the cylindrical curvature, this component is magnified by a factor $m$ ($x’_{\perp} = m \cdot x_{\perp}$).
Because one component is magnified while the other remains constant, the resultant vector (the image of the line) points in a new direction. This causes the lines to appear broken and tilted by an angle $\alpha$ relative to their original direction.
(b) Finding the Refractive Index ($\mu$)
Step 1: Calculate Transverse Magnification ($m$)
Consider the rod as a thick lens viewed from the top. The “object” (the ruled line) is in contact with the bottom surface of the rod. Let the radius of the rod be $R$.
- Object distance ($u$): The line is at the bottom of the diameter, so $u = -2R$ (sign convention: light travels up).
- Refraction at top surface: Light travels from glass ($\mu_1 = \mu$) to air ($\mu_2 = 1$). The radius of curvature for the top surface is $R$ (center of curvature is below, $C = -R$).
Using the single surface refraction formula $\frac{\mu_2}{v} – \frac{\mu_1}{u} = \frac{\mu_2 – \mu_1}{R_{surf}}$:
$$ \frac{1}{v} – \frac{\mu}{-2R} = \frac{1 – \mu}{-R} $$ $$ \frac{1}{v} = \frac{\mu – 1}{R} – \frac{\mu}{2R} = \frac{2\mu – 2 – \mu}{2R} = \frac{\mu – 2}{2R} $$ $$ v = \frac{2R}{\mu – 2} $$The magnification $m$ for a single spherical surface is given by $m = \frac{\mu_1 v}{\mu_2 u}$:
$$ m = \frac{\mu \cdot \left(\frac{2R}{\mu – 2}\right)}{1 \cdot (-2R)} = \frac{\mu}{2 – \mu} $$Step 2: Relate Magnification to Angles
Let the angle of the ruled lines with the rod’s axis be $\theta$. The slope of the line with respect to the axis is $\tan \theta = \frac{x_{\perp}}{y_{\parallel}}$.
The image line is tilted by an additional angle $\alpha$, so its total angle with the axis is $\theta + \alpha$. The new slope is:
Since $x’_{\perp} = m x_{\perp}$ and $y’_{\parallel} = y_{\parallel}$, we have:
$$ \tan(\theta + \alpha) = \frac{m x_{\perp}}{y_{\parallel}} = m \tan \theta $$ $$ m = \frac{\tan(\theta + \alpha)}{\tan \theta} $$Step 3: Solve for $\mu$
Equating the two expressions for $m$:
$$ \frac{\mu}{2 – \mu} = \frac{\tan(\theta + \alpha)}{\tan \theta} $$Rearranging to solve for $\mu$:
$$ \mu \tan \theta = (2 – \mu) \tan(\theta + \alpha) $$ $$ \mu \tan \theta = 2 \tan(\theta + \alpha) – \mu \tan(\theta + \alpha) $$ $$ \mu [\tan \theta + \tan(\theta + \alpha)] = 2 \tan(\theta + \alpha) $$