Question

Acceptance angle $\left(\theta \right)$ for an optical fiber is defined as the maximum angle of incidence at the interface of medium (air usually) and core $(\mu =n_1)$ for which light ray enters and travel along the optical fiber. Sine of acceptance angle is known as numerical aperture of the optical fiber. Then in the given setup, what is the value of numerical aperture?

• A. $n_1$
• B. $\frac{n_2}{n_1}$
• C. $\frac{n_1}{n_2}$
• D. $\sqrt{n_1^2-n_2^2}$

Answer 1

The correct option is D $\sqrt{n_1^2-n_2^2}$


Applying Snell's law for $\text{AB}$, ${\mu }_{air}\sin \theta =n_1\sin {\theta }_1$

$\Rightarrow \sin \theta =n_1\sin {\theta }_1[\because {\mu }_{air}=1]$

Applying snell's law for $\text{BC}$ (TIR condition) , $n_1\sin (90-{\theta }_1)=n_2$

$\cos {\theta }_1=\frac{n_2}{n_1}$

$\therefore \sin {\theta }_1=\frac{\sqrt{n_1^2-n_2^2}}{n_1}$

$\Rightarrow \sin \theta =\sqrt{n_1^2-n_2^2}$

Hence, option $\left(d\right)$ is correct