Determine: $i. The \ \ critical \ \ angle. \\ ii. The NA. \\ iii. The \ \ acceptance \ \ angle$

Given:
$n_1=1.5$

$n_2=1.47$

To Find:

1) Critical angle (Өc)

2) Numerical aperture (N.A)

3) Acceptance angle (Өa)

Solution:

• $θ_{c}=sin^{-1} \big(\frac{n_2}{n_1}\big)=sin^{-1} \big(\frac{1.47}{1.5}\big)$

• $θ_{c}$=78.52°

$\boxed {Critical \ \ angle ( θ_{c }) = 78.52°}$

• $N.A = \sqrt{n_1^2-n_2^2 }$

• $=\sqrt{1.5^2-1.47^2 }$

• N.A= 0.2984

$\boxed{Numerical\ \ aperture \ \ (N.A) =0.2984}$

• N.A = sin $Ө_a$

  0.2984=sin $Ө_a$

  $Ө_a$ = 17.36°

$\boxed{Acceptance \ \ angle \ \ (Ө_a) = 17.36°}$