Equivalent focal length of coaxial lens system

Equivalent focal length of coaxial lens system:

Consider two thin lenses L₁ and L₂ focal length f₁ and f₂ respectively. These are placed in air separated by a distance ‘a'. Let a light ray AB is incident on lens L₁, parallel to the principal axis at a height X above the axis. After refraction of this ray through lens L₁, it deviates through an angle δ₁ and the emergent ray ( in the absence of L₂ ), passes through second principal focus of L₁,
Which is at D. The ray AB is deviated through an angle δ₁ by the lens L₁. In figure A.

Let, O₁B = x ,      O₁O₂  = a,      O₂ C = y

F₁ and F₂  are focal points of the optical system.
P₁ and P₂  are principal points of the optical system.
δ₁  = tan δ₁  = O₁B  = X
                         O₁D    f₁

.•. Derivation produced by first lens = δ₁= x
                                                                            f₁

The emergent ray from the lens L₁ meets the lens L₂ at C which is at a  height y above the axis. After refraction through lens L₂ the emergent ray meets the axis at F₂ which is second focal point of the lens system. The lens L₂deviated the ray BC further through  δ₂.

Geometrical Optics and Interference

.•. Deviation produced by the second lens
     = δ₂ = y / f₂

When AB , the incident ray and CF₂, the emergent ray are produced, they  meet at E. Then a single convex lens placed at EP₂ having focal length P₂F₂ ( = f ) is equivalent focal length of the lens system.

.•. Deviation produced by equivalent lens
    = δ = EP₂  /  P₂F₂

  δ = x / f

Deviation produced by equivalent lens system is  the algebraic sum of deviation produced by individual lens.

.•. δ = δ₁+ δ₂

.•. x = x +  y 
    f     f₁    f₂      ......( 1 )

From figure A.
           
            y = HO₂ - HC
               = x - a. δ₁
                                 {since angle = arc / radius
                                   δ₁ =  HC/ BH = HC / a }
               = x - a. x /f₁

           y  = x (1- a/f₁ )    .....( 2 )

   Putting this value of y in equation (1)

    x = x +  y (1- a  )
    f     f₁    f₂       f₁

     1= 1 +  1 (1- a  )
    f     f₁    f₂       f₁

    1= 1 +  1  -     a        ..... (3)
    f     f₁    f₂      f₁f₂


Where f is the equivalent focal length of coaxial lens system.

a) power of lens :

Power of a lens is its ability to converge or diverge the rays incident on it. The power of lens is the reciprocal of its focal length in meters.

 Power of lens =              1                          
                              Focal length in meters

The unit of power of lens is diopter.

A convex lens of small focal length produces a large converging effect while a convex lens of large focal length produces a small converging effect. So power of a convex lens is taken as positive. A concave lens produces divergence. Therefore, power of cancave lens produces divergence. Therefore , power of concave lens is taken as negative.

b) power of coaxial lens system:

When two thin lenses of focal lengths  f₁ and f₂ are placed coaxially and separated by a distance "a" , then equivalent focal length f of lens system is given by



    1= 1 +  1  -     a     
    f     f₁    f₂      f₁f₂

But, power =                  1                           
                         Focal length( in meters )

.•. P = P₁ + P₂ - aP₁P₂

.•. where P is the equivalent power of coaxial lens system.

c) Position of cardinal point :

(i) Second principal point

Let  β = O₂P₂ = Distance of the second principal point from the second lens.

And ɑ = Distance of the first principal point from the first lens.

The calculate the value of β we have to produced as follows: from figure .A.

O₂P₂ = P₂F₂ - O₂F₂ = f - O₂F₂   .....(4)

Now , the triangles, ∆EP₂F₂ and ∆CO₂F₂ are similar Geometrical Optics and IntererencIn

.•. EP₂ = P₂F₂   i.e   x =     f .  
    CO₂    O₂F₂          y     O₂F₂

  O₂F₂ = y. f
              x

.•. putting the value of ‘y' from equation (2) we get ,


  O₂F₂  = x (1- a  )f = ( 1- a  )f
                        f₁               f₁
                        x

putting the value of  O₂F₂  in equation ..(4) we get,

 O₂F₂ = f - (1-a  )f = a . f
                      f₁        f₁

Since P₂ lies on the left side of lens L₂.

•.• O₂F₂ = - β
     
.•.          β  = - a. f 
                         f₁        .....(5)

So second principal point ( plane ) P₂ is at a distance "β " to the left from the second lens.

ii) First principal point :  To calculate the value of ɑ we have to consider rays parallel to principal axis from the right hand side as shown in figure B.

Proceedings in the similar manner as described above we get,

 O₁P₁ = ɑ = a.f
                      f₂       .....(6)

So, first principal point P₁ is at a distance "ɑ" to the right from the first lens.

Now, as  the medium on the two side of the lens system is same, the nodal points N₁ and N₂ coincide with P₁ and P₂.

iii ) Second focal point : The distance of the second focal point F₂ from the second lens L₂ will be O₂F₂ and is given by.

O₂F₂ = P₂F₂ - P₂O₂
         = f - (- O₂P₂ )
         = f + O₂P₂
         = f + β
         = f + (- af )
                      f₁
         = f + ( 1- af )
                         f₁

iv) First focal point : The distance of the first focal point F₁ from the first lens L₁ will be O₁F₁ and is given by

       O₁F₁ = P₁F₁ - P₁O₁
                = -f - (-O₁P₁ )
                = -f + O₁P₁
                = -f + ɑ
                = - f + af
                            f₂
                = - f ( 1- a)
                              f₂

Geometrical Optics and Interference

Let O = Positive of object
        I = Positive of image
       P₁ = First principal point
       P₂ = Second principal point
       u = O₁O= Object distance from the first lens.
        v =O₂I= Image distance from the second lens.
        U= P₁O= Object distance from the first principal point.
        V= P₂I= Image distance from the second principal point.
        ɑ= Distance of the first principal point from first lens.
        β= Distance of the second principal point from second lens.

From figure, ( using sign convention)
           
        -U = -u +ɑ
         U = u - ɑ
And. V= v -β

 For lens combination

  1 _ 1 = 1    and magnification m = V
  V   U    f                                               U