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# Equivalent focal length 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

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

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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