SEM-3 UNIT-1 OPTCS �Ch. 1 OPTICAL SYSTEM AND CARDINAL POINTS

Book : A TEXTBOOT OF OPTICS

BY DR. N. SUBRAHMANYAM

BRIJ LAL

DR. M. N. AVADHANULU

1. INTRODUCTION

Single lenses are rarely used for image formation, as they suffer from various defects. In optical instruments such as cameras, microscopes, telescopes etc., a collection of lenses are employed for forming images of objects. An optical system consists of a number of lenses placed apart, and having a common principal axis. The image formed by such a coaxial optical system is good and almost free of aberrations.

2. CARDINAL POINTS

In the case of refraction through a thin lens, the thickness of the lens has been neglected in calculating the various formulae. It is then immaterial from which points of the lens the distances are measured. But we cannot apply the above approximation for an optical system consisting of a combination of lenses (or for a thick lens). One way of calculating the position and size of the image formed by an optical system is to consider refraction at each surface of a lens successively, but the method proves to be more tedious.

In 1841, Gauss showed that any number of coaxial lenses could be treated as a single unit, without the necessity of treating the single surfaces of lenses separately. The lens makers’ formula can be applied to the system provided the distances are measured from two hypothetical parallel planes, fixed with reference to the refracting system. The points of intersection of these planes with the axis are called the principal points or Gauss points. In fact there are six points in all, which characterize an optical system. They are

1. Two focal points,
2. two principal points and
3. two nodal points.

These six points are known as cardinal points of an optical system. The planes passing through these points and which are perpendicular to the principal axis are known as cardinal planes. The cardinal points and cardinal planes are intrinsic properties of a particular optical system and determine the image forming properties of the system. If these are known, one can find the image of any object without making a detailed study of the passage of the rays through the system. It is not necessary to consider the refraction of the ray at the various surfaces.

We describe here how to locate the cardinal points and planes for a coaxial optical system.

2.1 PRINCIPAL POINTS AND PRINCIPAL PLANES

Fig. 1 (a)

Consider an optical system having its principal focal points F₁ and F₂ . A ray OA travelling parallel to the principal axis and incident at A is brought to focus at F₂ in the image space of the optical system as shown in Fig. 1 (a). The actual ray is refracted at each surface of the optical system and follows the path OABF₂.

If we extend the incident ray OA forward and the emergent ray BF₂ backward, they meet each other within the optical system at H₂. Now, we can describe the refraction of the incident ray OA in terms of a single refraction at a plane passing through H₂. A plane drawn through the point H₂ and perpendicular to the axis may be regarded as the surface at which refraction takes place. This plane is called the principal plane of the optical system.

Thus, the four consecutive deviations of the light rays caused by the four surfaces of the optical system are equivalent to a single refraction at H₂, taking place at the principal plane. We now define the principal plane of an optical system as the loci where we assume refraction to occur without reference as to where the refraction actually occurs.H₂P₂ is the principal plane in the image space and is called the second principal plane. The point P₂ at which the second principal plane intersects the axis, is called the second principal point.

Fig. 1 (b)

By adopting similar procedure, as shown in Fig. 1(b). We can locate the principal plane H₁P₁ and principal point P₁ in the object space.

Consider the ray F₁S passing through the first principal focus F₁ such that after refraction it emerges along QW parallel to the axis at the same height as that of the ray OA (see Fig 1 a). The rays F₁S and QW when produced intersect at H₁. A plane perpendicular to the axis and passing through H₁ is called the first Principal plane. The point of intersection, P₁, of the first principal plane with the axis is called the first principal point.

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Fig. 2

It is seen from Fig. 2 that two incident rays are directed towards H₁ and after refraction seem to come from H₂. Therefore, H₂ is the image of H₁. Thus, H₁ and H₂ are the conjugate points and the planes H₁P₁ and H₂P₂ are a pair of conjugate planes. It is also seen that H₂P₂ = H₁P₁

Hence, the lateral magnification of the planes is +1. Thus, the first and second principal planes are planes of unit magnification and are therefore called unit planes and the points P₁ and P₂ are called unit points.

Note – The principal planes are conceptual planes and do not have physical existence within the optical system.

2.1.1. Some Remarkable Features of Principal Planes

1. Even a complex optical system has only two principal planes.
2. Between H₁ and H₂ all rays are parallel to the principal axis.
3. The location of the principal planes is characteristic of a given optical system. Their positions do not change with the object and image distance used.
4. The principal planes are conjugate to each other. An object in the first principal plane is imaged in the second principal plane with unit magnification. Any ray directed towards a point on the first principal plane emerge from the lens as if it originated at the corresponding point (at the same distance above or below the axis) on the second principal plane. Hence, the name unit planes.
5. The principal points H₁ and H₂ provide a set of references from which several system parameters are measured.

2.2 FOCAL POINTS AND FOCAL PLANES

If a parallel beam of light travelling from the left to the right (in object space) is incident on the optical system, the beam comes together at a point, F₂ on the other side (in image space) of the optical system. The beam passes through the point F₂ whatever may be its path inside the system. The point, F₂, is called the second focal point of the optical system. A beam of light passing the point F₁ on the axis of the object side is rendered parallel to the axis after emergence through the optical system (Fig. 2). The point F₁ is called the first focal point.

We can now define the focal points as follows:

The first focal point is a point on the principal axis of optical system such that a beam of light passing through it is rendered parallel to the principal axis after refraction through the optical system.

The second focal point is a point on the principal axis of the optical system such that a beam of light travelling parallel to the principal axis of the optical system, after refraction through the system, passes through it.

The planes passing through the principal focal points F₁ and F₂ and perpendicular to the axis are called first focal plane and second focal plane respectively. The main property of the focal planes is that the rays starting from a point in the focal plane in the object space correspond to a set of conjugate parallel rays in the image space. Similarly, a set of parallel rays in the object space corresponds to a set of rays intersecting at a point in the focal plane in the image space.

The distance of the first focal point from the first principal point, i.e., F₁P₁ is called the first focal length, f₁, of the optical system and distance of the second focal point from the second principal point, F₂P₂, is called the Second focal length, f₂. f₁ and f₂ are also known as the focal lengths in object space and image space respectively.

When the medium is same on the two sides of the optical system f₁ = f₂ (numerically).

2.3. NODAL POINTS AND NODAL PLANES

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Nodal points are points on the principal axis of the optical system where light rays, without refraction, intersect the optic axis. In a thin lens the nodal point is the centre of the lens.

Light passing through the centre of a thin lens does not deviate. In an optical system the centre separates into two nodal points. The planes passing through the nodal points and perpendicular to the principal axis are called the nodal planes. Whereas the principal planes are planes where all refraction are assumed to occur, the nodal planes are planes where refraction does not take place. Fig. 1 represents an optical system with the help of its cardinal planes. It is seen from the Fig. 1 that a ray of light, AN₁, directed towards one of the nodal points, N₁, after refraction through the optical system, along N₁N₂, emerges out from the second nodal point, N₂, in a direction N₂R, parallel to the incident ray. The distances of the nodal points are measured from the focal points.

2.3.1 The nodal points are a pair of conjugate points on the axis having unit positive angular magnification

Let H₁P₁ and H₂P₂ be the first and second principal planes of an optical system. Let AF₁ and BF₂ be its first and second focal planes respectively (Refer to Fig. 1 earlier) consider a point A situated on the first focal plane. From A draw a ray AH₁ parallel to the axis. The conjugate ray will proceed from H₂, a point in the second principal plane such that H₂P₂ = H₁P₁ and will pass through the second focus.

Take another ray AT₁ parallel to the emergent ray H₂F₂ and incident on the first principal plane at T₁. It will emerge out from T₂, a point on the second principal plane such that T₂P₂ = T₁P₁, and will proceed parallel to the ray H₂F₂. The points of intersection of the incident ray AT₁ and the conjugate emergent ray T₂R with the axis give the positions of the nodal points. It is clear that the two points N₁ and N₂ are a pair of conjugate points and the incident ray AN₁ is parallel to the conjugate emergent ray T₂R.

Further

The ratio

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Therefore, the points N1 and N2 are a pair of conjugate points on the axis having unit positive angular magnification.

2.3.2 The distance between two nodal points is always equal to the distance between two principal points.

Referring to Fig. 1(earlier topic), we see that in the right angled Δ^les T₁P₁N₁ = T₂P₂N₂

T₁P₁ = T₂P₂

∠T₁N₁P₁ = ∠T₂N₂P₂= α

Therefore, the two Δ^les are congruent

• P₁N₁ = P₂N₂
• Adding N₁P₂ to both the sides, we get
• P₁N₁ + N₁P₂ = P₂N₂ +N₁P₂
• P₁P₂ = N₁N₂ -------------------(2)
• Thus, the distance between the principal points N₁ and N₂ is equal to the distance between the principal points P₁ and P₂.

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2.3.3. The nodal points N₁ and N₂ coincide with the principal points P₁ and P₂ respectively whenever the refractive indices on either side of the lens are the same.

Now consider the two right angled Δ^les AF₁N₁ and H₂P₂F₂ in Fig. 1 (earlier topic)

AF₁ = H₂P₂

∠AN₁F₁ = ∠H₂F₂P₂

• The two Δ^les are congruent.

F₁N₁ = P₂F₂

But F₁N₁ = F₁P₁ + P₁N₁

• F₁P₁ + P₁N₁ = P₂F₂
• P₁N₁ = P₂F₂ – F₁P₁

Also P₂F₂ = +f₂ and P₁F₁=-f₁

∴ P₁N₁=P₂N₂=(f₁+f₂)

As the medium is the same, say air, on both the sides of the system.

f₂= -f₁

∴ P₁N₁ = P₂N₂ = 0 ------------(3)

Thus, the principal points coincide with the nodal points when the optical system is situated in the same medium.

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3. CONSTRUCTION OF THE IMAGE USING CARDINAL POINTS

From the knowledge of the cardinal points of an optical system, the image corresponding to any object placed on the principal axis of the system can be constructed. It is not necessary to know the position and curvatures of the refracting surfaces or the nature of the intermediate media. Only knowledge of cardinal points and cardinal planes is sufficient.

FIG. 1

Let F₁, F₂ be the principal foci, P₁, P₂ the principal points and N₁, N₂ the nodal points of the optical system, shown in fig. 1. AB is a linear object on the axis. In order to find the image of the point A we make the following construction.

(1) A ray AH₁ is drawn parallel to the axis touching the first principal plane at H₁. The conjugate ray will processed from H₂ a point on the second principal plane such that H₂P₂ = H₁P₁ and will pass through the second principal focus F₂.

(2) A second ray AF₁K₁ is drawn passing through the first principal focus F₁ and touching the first principal plane at K₁. Its conjugate ray will proceed from K₂ such that K₂P₂ = K₁P₁ and it will be parallel to the axis.

(3) A third ray A₁T₁N₁ is drawn which is directed toward the first nodal point N₁. This ray passes after refraction through N₂ in a direction parallel to AN₁.

The point of intersection of any of the above two refracted rays will give the image of A.

Let it be A₁. From A₁, if a perpendicular is drawn on to the axis, it gives the image A₁ B₁ of the object AB.

4. NEWTON’S FORMULA

Referring to the Fig. 1, it is seen that Δ^les ABF₁ and F₁K₁P₁ are similar.

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This is the Newton’s formula. In the foregoing discussion, the distances of the image and the object have been measured from their respective foci. But it is sometimes convenient to measure the conjugate distances from the principal points.

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5. CARDINAL POINTS OF A COAXIAL SYSTEM OF TWO THIN LENSES

We determine the cardinal points of a coaxial optical system by assuming first that the object at infinity and then the case of object located on the principal axis at a certain distance from the system. We shall find that the computations would yield identical results in both the cases. We shall also observe that the results are identical with those obtained in the deviation method used in Chapter-4.

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5.1. OBJECT AT INFINITY

We now consider an object located at infinity, as shown in Fig. 1.

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Fig. 1

5.1.1. Focal Length of the system

Now, we can write the expression for the refraction taking place at the surface of first lens as follows.

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As u₁ = ∞, we obtain OG = f₁

The equation for the refraction at the second lens may be written as

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Which is the same as equ. 4.30. Because the location of the focal point F₂ is determined by QF₂, which is known from the equation, the position of the principal point P₁ is specified by the value of f calculated from the equ. (5).

5.1.2. Cardinal Points

1. Second Principal point: Let us say the second principal plane H₂P₂ is located at a distance of L₂P₂ = β from the second lens L₂. According to sign conversion β would be negative as it is measured toward the left of the lens.

QF₂ = f – (- β) = f + β

We can determine β using the equation for f into the above relation. Thus

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But from equ. , we have

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This is identical to

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(ii) First principal point: BY considering a ray of light parallel to the axis and incident on the second lens L₂ from the right side (see Fig. 2),

we can show that the distance of first principal plane, L₁P₁ = α form the first lens L₁ is given by

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This is the same as the result

Fig. 2

(iii) Second Focal Point: Referring the Fig. 1, the distance of the second focal point F₂ from the second lens L₂ is given by

(iv) First Focal Point: The distance of the first focal points F₁ from the first lens L₁ is given by

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(v) And (vi) Nodal Points: As the Optical system is considered to be located in air, P₁ and P₂ are also the positions of nodal points N₁ and N₂ respectively.

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5.2 AN AXIAL POINT-OBJECT

Let us consider a point-object O is placed at a distance u from the first lens and the final image is formed at I. The first image due to the first lens is formed at I’. Application of Gauss formula gives

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The image I’ acts as a virtual object for the second lens and the final image formed at I. The object distance for the second lens is v’- d.

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Fig. 1

Using the equ. (1) into equ. (2), we get

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Suppose the focal length of the equivalent lens is f and the reduced object-distance U = u - α and the reduced image-distance V = u - β. Here α represents the distance of the first lens surface from the first principal plane and β represents the distance of second surface of the second lens from the second principal plane. Then,

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Multiplying the above expression by(u - α) (v - β) f, we obtain

(u - α) f - (v - β) f = (u - α) (v - β)

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Simplifying the above expression and rearranging the terms, we get

u v + u (- β - f) + v ( -α + f) + (α β - β f +α f) = 0 ----(7)

Comparing Equs. (4) and (7), we have

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On Simplifying the above equation we get

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A lens combination is characterized by two principal planes, namely first and second principal planes. α gives the position of the first principal plane; as it is a positive quantity, the plane is to the right of the first lens.

From equ. (8), we see that

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β Gives the position of the second principal plane; as it is a negative quantity, the plane is to the left of the first lens. The Position of focal points are given by expressions (6) and (7) (earlier topic).

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EX-1 A Coaxial lens system placed in air has two lenses of focal lengths 3F and F separated by a distance 2F. Find the positions of the cardinal points.

Solution: f₁ = 3F, f₂ = F, d=2F

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Therefore, the first principal point P₁ is at a distance 3F to the right of the first lens, as shown in Fig.1

Fig. 1

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The Second principal point P₂ is at a distance F to the left of the second lens.

The first focal point F₁ is at a distance 3F – 3F/2 =3F/2 from the first lens and F₂ is at a distance 3F/2 – F = F/2 from the second lens. As the medium on the two sides of the lens system is the same, the nodal points N₁ and N₂ coincide with P₁ and P₂.

EX-2 Two thin convex lenses of focal lengths 20 cm and 5 cm are kept coaxially separated by a distance of 10 cm. Plot the positions of the cardinal points for the combination.

Solution: Given that f₁=20 cm, f₂=5 cm and d=10cm

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Fig. 1

Nodal Points: As the system is situated in air, the nodal points are same as the principal points.

First Focal Point:

The distance of F₁ from the lens L₁ is

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Second Focal Points:

The distance of F₂ from the lens L₂ is

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The first principal point P₁ is to the right of the first lens and is at a distance of 13.33 cm from it. The second principal point P₂ is to the left of the second lens and is at a distance of 3.33 cm from it. The cardinal points are plotted in Fig. 1.

Ex-3 Two identical thin convex lenses of focal lengths 8 cm each are coaxial and 4 cm apart. Find the equivalent focal length and the positions of the principal points. Also, find the position of the object for which the image is formed at infinity.

Solution: Here, f1 = f2 = 8 cm, d = 4 cm

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The points are plotted in Fig. 1. For the final image to be formed at infinity.

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Fig. 1

As V = ∞ (given), f = 5.33 cm, we obtain U = -f = -5.33 cm. But U=u - α.

• u = U + α = -5.33 cm + (+2.66 cm) = -2.67 cm.
• Therefore, the object is at a distance of 2.67 cm to the left of the first lens.

Ex-4: Two thin convergent lenses each of 20 cm focal length are set coaxially 5 cm apart. An image of upright pole 200 m distant and 10 m high is formed by the combination. Find the position of the unit and focal planes and the image. Also find the size of the image.

Solution: Here, f₁ = f₂ = 20 cm, d = 5 cm

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We have u = - 20000 cm, α = 2.85 cm.

U=u - α = -20000 cm – 2.85 cm = - 20002.85 cm

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Or V = 11.4 cm

But V = v - β

• v = V + β = 11.4 cm – 2.85 cm = 8.55 cm

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Therefore, the first unit plane P₁ is at a distance of 2.85 cm to the right of the lens L₁ and second unit plane P₂ is at a distance of 2.85 cm to the left of the lens L₂.

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The first focal plane is at a distance of 11.43 cm – 2.85 cm = 8.58 cm to the left of the lens L1 and the second focal plane is at a distance of 8.58 cm to the right of the lens L2.

The final image is formed at a distance of 8.55 cm to the right of the lens L2.

Magnification,

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Fig. 1

Ex-5 : A thin convex and a thin concave lens each of 50 cm focal length are coaxially situated and separated by 10 cm. Find the position and nature of the final image formed of an object placed 20 cm from the convex lens (i.e. 30 cm from the concave lens).

Solution : Here, f₁ = + 50 cm, f₂ = - 50 cm, d = 10 cm

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U = 30 cm, f = +250 cm.

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The image is formed at a distance 26.8 cm from P2. Therefore, the image is at a distance 13.21 cm to the left of the convex lens. The image is virtual and erect.

Fig. 1

Ex-6 : An optical system consists of two thin lenses, a convex lens of focal length 20 cm and a concave lens of focal length 10 cm and separated by a distance of 8 cm. An object 1 cm in height is placed at a distance 50 cm from the convex lens. Find the position and size of the image.

Solution: Here, f1 = + 20 cm, f2 = - 10 cm, d = 8 cm

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The first principal point P1 is at a distance 80 cm to the right of the convex lens.

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Fig. 1

The second principal point P2 is at a distance of 40 cm to the right side of the concave lens.

For the combined system, distance of the object from P1