OPTICS OBJECTIVE 15

Problem 15: Imaging with Negative Refractive Index Slab

Step-by-Step Analysis

We analyze the refraction at each surface using the formula $\frac{n_2}{v} – \frac{n_1}{u} = 0$ (for plane surfaces, $R=\infty$).

1. Refraction at the First Surface (Left Face)

• Medium 1: Air ($n_1 = 1$), Medium 2: Slab ($n_2 = -1$).
• Object: Real object at distance $x_o$. So $u_1 = -x_o$.
• Formula: $\frac{-1}{v_1} = \frac{1}{-x_o} \implies v_1 = x_o$.
• The rays bend to the same side of the normal. Diverging rays from the object converge to a point inside the slab at distance $x_o$ from the first face. Let’s call this point $I_1$.

2. Refraction at the Second Surface (Right Face)

The image $I_1$ acts as the object for the second refraction. The second face is at distance $d$.

Case A: $x_o \le d$

• The convergence point $I_1$ is inside the slab (or on the back face).
• Rays diverge from $I_1$ towards the second face. Thus, for the second face, we have a Real Object.
• Object distance relative to second face: $u_2 = -(d – x_o)$.
• Refraction ($n_1 = -1 \to n_2 = 1$): $$ \frac{1}{v_2} = \frac{-1}{-(d – x_o)} = \frac{1}{d – x_o} \implies v_2 = d – x_o $$
• Since $x_o \le d$, $v_2 \ge 0$. Rays converge to form a Real Image outside the slab.

Case B: $x_o > d$

• The convergence point $I_1$ is beyond the second face.
• Rays hit the second face while still converging. Thus, for the second face, we have a Virtual Object.
• Object distance (to the right): $u_2 = +(x_o – d)$.
• Refraction: $$ \frac{1}{v_2} = \frac{-1}{+(x_o – d)} \implies v_2 = -(x_o – d) $$
• Since $v_2$ is negative, the rays diverge after exiting the slab. They appear to originate from a point to the left of the face, forming a Virtual Image.