(a) Let us first derive the condition for total internal reflection.
Critical angle for the interface of medium 1 and medium 2.
sin C =

1

1 ‌ µ2

=

µ2

µ1

=

1.44

1.68

= 0.8571
So, critical angle C = 59°
Condition for total internal reflection from core to cladding
i2 > 59° or r ≤

π

2

- 59° or r ≤ 31°
Now, for refraction at first surface air to core.
Snell’s law,

sini1

sinr

=

a

‌

µ1
sin i1 =

a

‌

µ1 sin r = 1.68 sin 31° or i1 = 60°
Thus all incident rays which makes angle of incidence between 0° and 60° will suffer total internal reflection in the optical fibre.
(b) When there is no outer covering critical angle from core to surface.
sin C =

1

1 ‌ µa

=

µa

µ1

sinC =

1

1.68

⇒ C = sin−1(

1

1.68

) = 36.5°
So, condition for total internal reflection from core to surface
i2 > 36.5° or r <

π

2

- 36.5° or r < 53.5°
Let us find range of incident angle at first surface air to core.

a

‌

µ1 =

sini1

sinr

sin i1 = 1.68 × sin 53.5° = 1.68 × 0.8039
sin i1 = 1.35 or i ≈ 90°.
Thus all incident rays at first surface 0° to 90° will suffer total internal reflection inside core.