OPTICS OBJECTIVE 12,13,14

Anamorphic De-magnifier Prism

12. Refractive Index Expression

Using the geometry shown in the diagram:

1. Incidence (Vertical Face): Normal incidence ($i=0$), travels horizontally inside.
2. Reflection 1 (Hypotenuse): Incidence angle $i = 90^\circ – \theta$. Reflected ray deviates by $2(180-i)$ relative to incident path, or geometrically, it makes an angle $2\theta$ downward with the horizontal.
3. Reflection 2 (Base): Incident at $2\theta$ to the horizontal base. Reflects at $2\theta$ upwards.
4. Emergence (Hypotenuse):
   • Ray travels at angle $2\theta$ upward.
   • Normal to hypotenuse is at $90^\circ – \theta$ to the horizontal.
   • Angle of incidence (inside): The angle between the ray ($2\theta$) and the normal ($90^\circ – \theta$).
     $i’ = (90^\circ – \theta) – 2\theta = 90^\circ – 3\theta$.
   • Angle of emergence (outside): The ray emerges horizontally ($0^\circ$). The angle with the normal is:
     $e = 90^\circ – \theta$.

Applying Snell’s Law ($ \mu \sin i’ = 1 \sin e $):

$$ \mu \sin(90^\circ – 3\theta) = \sin(90^\circ – \theta) $$ $$ \mu \cos(3\theta) = \cos(\theta) \implies \mu = \frac{\cos\theta}{\cos 3\theta} $$

13. Range of $\theta$

Conditions for the device to work:

• TIR at Base: Angle with normal is $90^\circ – 2\theta$. This must exceed critical angle $C$. $$ 90^\circ – 2\theta > C \implies \sin(90^\circ – 2\theta) > \sin C $$ $$ \cos 2\theta > \frac{1}{\mu} \implies 2\theta < \cos^{-1}\left(\frac{1}{\mu}\right) \implies \theta < \frac{1}{2}\cos^{-1}\left(\frac{1}{\mu}\right) $$
• Transmission at Exit: Angle of incidence $90^\circ – 3\theta$ must be less than critical angle $C$. $$ 90^\circ – 3\theta < C \implies \sin(90^\circ - 3\theta) < \frac{1}{\mu} $$ $$ \cos 3\theta < \frac{1}{\mu} \implies 3\theta > \cos^{-1}\left(\frac{1}{\mu}\right) \implies \theta > \frac{1}{3}\cos^{-1}\left(\frac{1}{\mu}\right) $$

14. Derivation of Vertical De-magnification Ratio

The “vertical de-magnification ratio” is the ratio of the beam widths perpendicular to the direction of propagation.

Expressing in terms of $\mu$
We established in Q12 that $\mu = \frac{\cos\theta}{\cos 3\theta}$. We need to transform $\frac{\sin\theta}{\sin 3\theta}$.

Using identities $\cos 3\theta = 4\cos^3\theta – 3\cos\theta$ and $\sin 3\theta = 3\sin\theta – 4\sin^3\theta$:

$$ \mu = \frac{\cos\theta}{4\cos^3\theta – 3\cos\theta} = \frac{1}{4\cos^2\theta – 3} $$ $$ 4\cos^2\theta – 3 = \frac{1}{\mu} \implies 4(1-\sin^2\theta) – 3 = \frac{1}{\mu} $$ $$ 1 – 4\sin^2\theta = \frac{1}{\mu} \implies 4\sin^2\theta = 1 – \frac{1}{\mu} = \frac{\mu-1}{\mu} $$

Now look at the ratio $R = \frac{\sin\theta}{\sin 3\theta} = \frac{\sin\theta}{3\sin\theta – 4\sin^3\theta} = \frac{1}{3 – 4\sin^2\theta}$.

Substitute $4\sin^2\theta$ from above:

$$ R = \frac{1}{3 – (\frac{\mu-1}{\mu})} = \frac{\mu}{3\mu – (\mu – 1)} = \frac{\mu}{2\mu + 1} $$