The equivalent focal length f for two thin lenses separated by distance x is given by

In many optical instruments there may be compound lenses, that is, two or more lenses in contact. We will first deal with the case of two thin lenses in contact.

In Figure 1, let the be focal lengths of the two lenses be f1 and f2.
u1 = OC1 and v1 = I1C1
u2= -C1I1 which is approximately equal to C2 I1 and v2 = C2I which is approximately equal to C1ITherefore

1/u1 + 1/v1 = 1/f1 and 1/u2 + 1/v2 = 1/f2

1/OC1 + 1/I1C1 = 1/f1 and 1/- C1I1 + 1/C1I = 1/f2Therefore:

1/OC1 + 1/I1C1 = 1/f1 + 1/f2 = 1/F

Combined focal length of two thin lenses in contact is given by:

1/F = 1/f1 + 1/f2

Combined focal length (F): =f1f2/[f1 + f2]

where F is the focal length of the combination.

Example problem A bi-convex lens of focal length 10 cm is fixed to a plano concave lens of focal length 20 cm made of glass of the same refractive index. What is the focal length of the combination?

Combined focal length (F): =f1f2/[f1 + f2]

F = 10x(-20)/[10-20] = +20 cm The focal length of the combination is positive and so it acts as a convex lens.

The combined focal length for two thin lenses separated by a distance a (Figure 2) is given by the equation:

1/F = 1/f1 + 1/f2 - a/f1f2

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Hint: Focal length is the distance over which the parallel rays either converge or diverge. For convex lenses, the focal length is positive and for concave length it is negative. Focal length varies with sign depending on the nature of the lens and the mirror.

Complete solution:

Let us consider two lenses A and B of focal length $f_1$ and $f_2$ which are placed in contact with each other. The object is placed at a point O beyond the focus of the first lens A. $I_1$ is the image produced by the first lens which is real. Since it is a real image, it serves as a virtual object for the second lens B producing the final image at $I$. The direction of the rays emerging from the first lens gets changed according to the angle at which strikes the second lens. We assume that the optical centers of the lenses are coincident as the lenses are very thin. Let this central point be represented by P. Therefore, For the image formed by lens A, we get$\dfrac{1}{{{v_1}}} - \dfrac{1}{u} = \dfrac{1}{{{f_1}}}\_\_\_\_\_\_\_\left( 1 \right)$Similarly, for the image formed by lens B, we get$\dfrac{1}{v} - \dfrac{1}{{{v_1}}} = \dfrac{1}{{{f_2}}}\_\_\_\_\_\_\_\left( 2 \right)$On adding $($1 ) and $($2 ), we get\[\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_2}}}\]If the two lenses are considered as equal to a single lens of focal length f, we get$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$\[ \Rightarrow \dfrac{1}{f} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_2}}}\]\[ \Rightarrow \dfrac{1}{f} = \dfrac{{{f_2} + {f_1}}}{{{f_1}{f_2}}}\]\[ \Rightarrow f = \dfrac{{{f_1}{f_2}}}{{{f_1} + {f_2}}}\]

The expression is \[f = \dfrac{{{f_1}{f_2}}}{{{f_1} + {f_2}}}\].

Note: 1) The focal length of the mirror is the distance between the pole and the focus of the mirror.

2) Focal length of the mirror is a measure of the power of the mirror.3) Thin lenses have the same focal length on either side.

Fig. 1. Combination of two Lenses separated by a distance.

Equivalent focal length of two lenses :- Let us consider two lenses L₁ ( focal length f₁ ) and L₂ ( focal length f₂ ) are separated by a distance d and P₂ is equivalent lens of both lenses ( focal length f ). Angle of deviation produce by the lenses P, L₁ and L₂ are δ, δ₁ and δ₂ respectively, then from the above figure.

δ = δ₁ + δ₂

h₁/f = h₁/f₁+ h₂/f₂ ………….( 1)

From ∆ AL₁F₂’ and ∆ BL₂F₂’

AL₁/L₁F₂’ = BL₂/L₂F₂’

AL₁/L₁F₂’ = BL₂/L₁F₂- L₁L₂

h₁/f₁ = h₂/f₁ – d

Hence, h₂ = h₁( f₁- d )/f₁

Put the value of h₂ in equation ( 1 )

h₁/f = h₁/f₁ + h ( f₁ – d )/f₁f₂

1/f = 1/f₁ + ( f₁ – d )/f₁f₂

1/f = 1/f₁ + 1/f₂ – d/f₁f₂

Hence, f = f₁f₂/f₁ + f₂ – d …………….. ( 2 )

Position of equivalent lens :-

From ∆ CP₂F₂ and ∆ BL₂F₂

P₂F₂/L₂F₂ = CP₂/BL₂

f/f – x₂ = h₁/h₂

Here P₂F₂ = f

According to the sign convention

f/f – ( – x₂ ) = h₁/h₂

f/f + x₂ = h₁/h₂

since, h₂ = h₁ ( f₁ – d )/f₁

hence, f/f + x₂ = f₁/f₁ – d

or ff₁ – fd = ff₁ + x₂f₁

or x₂ = – fd/f₁

From equation ( 2 )

or x₂ = – f₁f₂d/f₁ + f₂ – d f₁

or x₂ = -f₂d/f₁ + f₂ – d

Where x₂ is the position of equivalent lens.

Power of equivalent lens : – If the Power of first and second lenses are P₁ and P₂ respectively, then the Power of equivalent lens

P = P₁ + P₂ – dP₁P₂

Also read:- Deviation without Dispersion And Dispersion without deviation

Angle of Deviation , Refractive index of the Prism

Refraction of light at a spherical surface

Question 1. Find the position of the image formed by the lens combination given in the the figure.

Fig. 2. Combination of Lenses.

Solution.

Image formed by the first lens.

1/f₁ = 1/v₁ – 1/u₁

1/v₁ = 1/f₁ + 1/u₁ = 1/10 – 1/30 = ( 3 – 1 )/30 = 1/15

or v₁ = 15 cm

Image formed by the first lens serves as the object for the second lens. This is at ditance 15 – 5 = 10 cm for second lens.

f₂ = – 10

u₂ = 10

v₂ = ?

1/f₂ = 1/v₂ – 1/u₂

1/v₂ = 1/f₂ + 1/u₂ = 1/- 10 + 1/10 = 0/10

v₂ = infinite ( ∞ )

For the third lens.

f₃ = 30 cm

u₃ = ∞

v₃ = ?

1/v₃ = 1/f₃ + 1/u₃ = 1/30 – 1/∞

v₃ = 30 cm.

The final image is formed at 30 cm to the right of the third lens.

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I have found the formula for the effective focal length $f$ of two thin lenses with focal lengths $f_1$ and $f_2$ separated by distance $d$ to be $$ \frac 1f=\frac 1{f_1}+\frac 1{f_2}-\frac d{f_1f_2}. $$ However, I can't seem to find how $f$ is defined. Is it the distance from the first lens to the final focal point or the distance from the second lens to the final focal point? Or neither?