Question

derive the relation 1/f=1/f1+1/f2-d/f1f2​

Answer 1

Consider two lenses L1 and L2 separated by a small distance 'd' apart, as shown below. A ray of light AB initially parallel to the principal axis hits the lens L1 and deviates and then hits lens L2.

here

f1 is the focal length of L1 and

f2 is the focal length of L2

and

δ1 is the deviation produced by L1

δ2 is the deviation produced by L2

so,

form simple geometry

δ1 = h1 / F1

and

δ2 = h2 / F2

now,  

total deviation of the light ray will be

δ = δ1  +δ2

or

δ =  (h1 / F1) +  (h2 / F2)

also

at lens 2 and triangle BCD

h2 = h1 - CD = h1 - BD.tanδ1

or

h2 = h1 - d.tanδ1

now as δ1 is very small, tanδ1 ~ δ1

thus,

h2 = h1 - d.δ1

so, from earlier relation

h2 = h1 - d.(h1/f1)

thus,

δ = δ1  +δ2 = (h1/f1) +  [(h1 - d.(h1/f1)) / f2]

or

δ =  (h1/f1) + (h1/f2) - (dh1 / f1.f2)

now,

for the combination of the two lenses let F be the combined focal length.

So, the total deviation will be given as  

δ =  h1 / F

or

(h1/f1) + (h1/f2) - (d.h1 / f1.f2) = h1 / F

thus,

1/F = 1/f1 + 1/f2 - (d / f1.f2)

so, the combined focal length will be

F = f1.f2 / (f1 + f2 - d)

Answer 2

Answer:

The derivation of the relation 1/f=1/f1+1/f2-d/f1f2​1/f = 1/f1 + 1/f2 - 1/d

Explanation:

From the above question,

They have given :

Derive the relation :

$(1)/(F) =(1)/(f_(1)) +(1)/(f_(2))$

where f1  and f2  are focal length of two thin lenses and F is the focal length of the combination in contact.

Let us consider two thin lenses of focal lengths f1 and f2, separated by a distance d.

Let us consider an object of distance x1 from the first lens, and a final image at a distance x2 from the second lens.

According to the thin lens equation,

1/f1 = (1/x1) - (1/x2) and  1/f2 = (1/x2) - (1/x3)

where x3 is the distance from the second lens to the final image.

Adding the two equations, we get

1/f1 + 1/f2 = (1/x1) - (1/x3)

Substituting x3 = x2 + d, we get

1/f1 + 1/f2 = (1/x1) - (1/x2) - (1/x2 + d)

Simplifying the equation, we get

1/f1 + 1/f2 = 1/x1 - (1/x2 + 1/d)

Rearranging the terms, we get

1/f = 1/f1 + 1/f2 - 1/d

This is the required relation.

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