# Acceptance angle and Numerical Aperture

In this lecture, we are going to learn about the definition of Acceptance angle, then we will derive the formula for Acceptance angle and numerical aperture. So let’s start with the Definition of the Acceptance angle.

## Definition of Acceptance Angle in Optical Fiber

The acceptance angle of the fiber \phi_{max} is defined as the maximum value of the angle of incidence at the entrance (air/fiber interface) end of the fiber, at which the angle of incidence at the core-cladding interface is equal to the critical angle of the core medium.

Mathematically given as,

\boxed{\sin \Phi_{max}= \sqrt{n_1^2-n_2^2}}

where n1 and n2 are the refractive indexes of the core and the cladding respectively.

## Derivation of Acceptance Angle in Optical Fiber

The above mathematical relation can be obtained as follows. The below figure shows the longitudinal cross-section of the launch end of the fiber with a ray entering it.

The core of the fiber has a refractive index of n1 and is surrounded by a cladding of a material with a lower refractive index of n2.

• Light is launched into the end of the fiber from a launch region with a refractive index of n0. If the launch region is air then n0=1. When light rays enter the fiber, they strike the air/glass interface at its axis point A, with an angle of incidence \phi_{in}.
• Consequently, the light entering the air/glass interface propagates from a less dense medium into a more dense medium. Under these conditions and according to Snell\s law, the light rays will refract towards the normal (i.e. normal B).
• The angle of refraction is denoted as \phi_1.
• It is then reflected from the core/cladding interface at point B with an internal incidence angle \phi_c. [This is different from the external angle of incidence at the air/glass interface \phi_{in}.]
• In order for a ray of light to propagate down the cable (i.e. for total internal reflection to take place), it must strike the internal core/cladding interface at an angle that is greater than the critical angle \phi_c.
• Applying Snell’s law to the external angle of incidence yields, the following relation

n_0\sin\Phi_{in}=n_1\sin\Phi_1 ……(1)

But, \Phi_1=90^o-\Phi_c ……(2)

• Therefore, substituting equation (2) in (1) we get,

n_0 \sin\phi_{in}=n_1\sin(90^o-\phi_c)

n_0 \sin\phi_{in}=n_1\cos\phi_c \; [\because \sin(90^o-\phi_c)=\cos\phi_c] ……(3)

• If we consider point B in the above figure, the critical angle value for \phi_c is

\sin\phi_c=\frac{n_2}{n_1} ……(4)

• Expressing the term \cos\phi_c, in equation (3), in terms of \sin\phi_c, then equation (3) modifies to the form,

n_0\sin\phi_{in}=n_1(1-\sin^2\phi_c)^{1/2} ……(5)

• Substituting for\sin\phi_c from equation (4)

n_0\sin\phi_{in}=n_1 \left [1-\left (\frac{n_2}{n_1}\right)^2 \right]^{1/2}

n_0\sin\phi_{in}=n_1 \left (\frac{n_1^2-n_2^2}{n_1^2} \right)^{1/2}

or n_0\sin\phi_{in}=\frac{n_1}{n_1} \sqrt{n_1^2-n_2^2} ……(6)

• Because light rays generally enter the fiber from the air medium. no=1. Therefore, the maximum value of \sin\phi_{in} is given as,

\boxed{\sin\Phi_{in(max)}=\sqrt{n_1^2-n_2^2}}

or \boxed{\Phi_{in(max)}=\sin^{-1}\sqrt{n_1^2-n_2^2}} ……(7)

• Thus, \phi_{in} is called the acceptance angle or acceptance cone half-angle.

Thus,

1. The acceptance angle of the fiber is the maximum angle at which a light ray can enter into the fiber and still be totally internally reflected.
2. A cone of light incident at the entrance end of the fiber will be guided through the fiber, provided the semi-vertical angle of the core is less than or equal to \phi_{in(max)}.
Note: The angle \phi_{in(max)} is unique only for a particular fiber. it differed from fiber to fiber and depends on the material and the core diameter.

## Derivation of Numerical Aperture of Optical Fiber

• Numerical Aperture (NA) is a Figure of merit that is used to describe the light-gathering or light-collecting ability of an optical fiber.
• The larger the magnitude of NA, the greater the amount of light accepted by the fiber from the external light source.
• Numerical Aperture is mathematically defined as the sine of the acceptance cone half-angle.

Thus,

NA=\sin\Phi_{in(max)}

or NA=\sqrt{n_1^2-n_2^2} ……(8)

• The NA can also be expressed in terms of normalized refractive index difference or relative index(\Delta).
• It is defined as the ratio of the refractive index difference between the core and cladding to the refractive index of the core.

i.e., \Delta=\frac{n_1^2-n_2^2}{2n_1^2} =\frac{n_1-n_2}{n_1} for all practical purpose.

\therefore NA=\sqrt{n_1^2-n_2^2}=\sqrt{(n_1+n_2)(n_1-n_2)}

=\sqrt{(n_1+n_2)n_1\Delta} \;\;\left[\because \frac{n_1-n_2}{n_1}=\Delta\right]

=\sqrt{2n_1^2\Delta} \;\;if\; n_1=n_2

\boxed{\therefore NA =n_1\sqrt{2\Delta}}

• It can be noted that the NA is effectively dependent only on the refractive indices of the core and cladding material and is not a function of the fiber dimension.

## Frequently Asked Questions on Acceptance angle and Numerical Aperture

1. ### What is meant by the acceptance angle?

The acceptance angle is the maximum angle at which incoming sunlight can be captured by a solar concentrator. Its value depends on the concentration of the optic and the refractive index in which the receiver is immersed.

2. ### What does numerical aperture do?

The numerical aperture of a microscope objective is the measure of its ability to gather light and to resolve fine specimen detail while working at a fixed object (or specimen) distance.

3. ### Why is numerical aperture important?

The numerical aperture (abbreviated as ‘NA’) is an important consideration when trying to distinguish detail in a specimen viewed down the microscope. NA is a number without units and is related to the angles of light which are collected by a lens.

4. ### What are the advantages of a high numerical aperture objective?

The larger the NA, the stronger the ability of optical fiber to receive light. From the point of view of increasing the optical power entering the optical fiber, the NA is larger for better, because the larger the numerical aperture of the optical fiber is, the better the docking of the optical fiber.

Hello friends, my name is Trupal Bhavsar, I am the Writer and Founder of this blog. I am Electronics Engineer(2014 pass out), Currently working as Junior Telecom Officer(B.S.N.L.) also I do Project Development, PCB designing and Teaching of Electronics Subjects.

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