OPTICS OBJECTIVE 16

Problem 16: Virtual Object and Negative Index Slab

Step-by-Step Analysis

A “virtual object at distance $x_o$” means incident rays are converging towards a point distance $x_o$ behind the first surface (inside the slab geometry).

1. Refraction at First Surface (Face A)

• Object: Virtual ($u_1 = +x_o$).
• Refraction: $\frac{-1}{v_1} = \frac{1}{x_o} \implies v_1 = -x_o$.
• The negative sign indicates the image is formed at distance $x_o$ to the left of Face A (in front of the slab).
• Physically, the converging incident rays refract to become diverging rays that appear to originate from a point $P’$ in front of the slab.
• Lateral Magnification $m_1 = \frac{n_1 v_1}{n_2 u_1} = \frac{(1)(-x_o)}{(-1)(x_o)} = +1$. The intermediate image is Erect.

2. Refraction at Second Surface (Face B)

• The diverging rays from $P’$ travel through the slab thickness $d$.
• Object: Real object at distance $u_2 = -(x_o + d)$ from Face B.
• Refraction: $\frac{1}{v_2} = \frac{-1}{-(x_o + d)} \implies v_2 = x_o + d$.
• Since $v_2 > 0$, the rays converge to form a Real Image at distance $x_o + d$ outside the slab.
• Lateral Magnification $m_2 = \frac{n_2 v_2}{n_3 u_2} = \frac{(-1)(x_o + d)}{(1)(-(x_o + d))} = +1$.

Conclusion

Total Magnification $M = m_1 \times m_2 = 1 \times 1 = 1$. The final image is Real and Erect. This result holds regardless of whether $x_o < d$ or $x_o > d$.