OPTICS OBJECTIVE 17

Problem 17: Relationship Graph between Object and Image Distance

Analysis using Cartesian Coordinates

We are plotting the final image coordinate $x_i$ versus the object coordinate $x_o$, both measured from the front face A.

• Coordinate System: Origin at Face A. Light travels in $+x$ direction.
  • The real object is placed in front of the slab, so its coordinate $x_o$ is negative ($x_o < 0$).
  • The slab extends from $x=0$ to $x=d$.

Derivation of the Relationship

Let’s use the results from Problem 15, applying strict sign conventions.

1. Refraction at A ($x=0$):
   • Object position: $x_o$ (negative).
   • Equation: $\frac{n_2}{v_1} = \frac{n_1}{u_1}$. Here $u_1 = x_o$.
   • $\frac{-1}{v_1} = \frac{1}{x_o} \implies v_1 = -x_o$.
   • The intermediate image position is $x_1 = -x_o$ (positive, inside the slab).
2. Refraction at B ($x=d$):
   • Object position relative to B: The object is at $x_1 = -x_o$. Face B is at $d$.
   • Object distance $u_2 = \text{Position of object} – \text{Position of pole} = (-x_o) – d$.
   • Equation: $\frac{n_{out}}{v_{2\_rel}} = \frac{n_{slab}}{u_2}$.
   • $\frac{1}{v_{2\_rel}} = \frac{-1}{-x_o – d} = \frac{1}{x_o + d}$.
   • $v_{2\_rel} = x_o + d$. (This is the distance relative to pole B).
   • Absolute position $x_i = \text{Position of pole B} + v_{2\_rel} = d + (x_o + d)$.

Graph Analysis

The relationship is linear: $y = x + c$ with slope $+1$ and intercept $+2d$.

• At $x_o = -2d$, $x_i = 0$.
• At $x_o = 0$, $x_i = 2d$.

Graph (b) shows a single straight line with slope 1, passing through $(-2d, 0)$ and $(0, 2d)$. This matches our derived equation exactly.