Question

what is the numerical aperture if refractive index is 2.33 and critical angle is 350​

Answer 1

Answer:

Sol: Let the refractive index of core, n1 = 1.48

and the refractive index of cladding , n2 = 1.45

Numerical aperture (NA) images

images

Let θ0 be the acceptance angle

n2 sin 90 = n1 sin θc [θc = critical angle]

The fractional refractive ...

Answer 2

Answer:

The numerical aperture is 230.

Explanation:

The numerical aperture is given as,

$N=\mu cos\theta_c$  (1)

Where,

N=the numerical aperture of the object

μ=refractive index of the medium

θc=critical angle of the medium

From the question we have,

μ=2.33

θc=350

By substituting the required values in equation (1) we get;

$N=2.33\times cos 350$

$N=2.33\times 0.9848$          (cos350=0.9848)

$N=2.294584\approx230$

Hence, the numerical aperture is 230.