Two thin convex lenses of focal lengths f1 and f2 are separated by a horizontal distance d (where d < f1 < f2) and their centres are displaced by a vertical separation D as shown in the figure. Taking the origin O, at the centre of first lens, the x and y co-ordinate of the focal point of this lens system, for a parallel beam of rays coming from the left, are given by

Engineering

Physics

Curved Surface Refraction

Question

Two thin convex lenses of focal lengths f₁ and f₂ are separated by a horizontal distance d (where d < f₁ < f₂) and their centres are displaced by a vertical separation D as shown in the figure. Taking the origin O, at the centre of first lens, the x and y co-ordinate of the focal point of this lens system, for a parallel beam of rays coming from the left, are given by

$x = \frac{f_{1} (f_{1} + d)}{f_{1} + f_{2} - d}, y = \frac{Δ}{f_{1} + f_{2}}$

$x = \frac{f_{1} f_{2}}{f_{1} + f_{2}}, y = Δ$

$x = \frac{f_{1} f_{2} + d (f_{1} - d)}{f_{1} + f_{2} - d}, y = \frac{Δ (f_{1} - d)}{f_{1} + f_{2} - d}$

$x= \frac{f_{1} f_{2} + d (f_{1} - d)}{f_{1} + f_{2} - d}, y = \frac{{Δf}_{2}}{f_{1} + f_{2} - d}$

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Lens -1

$\frac{1}{v'} - \frac{1}{u} = \frac{1}{f_{1}}$

$u = \infty$

$\frac{1}{v'} = \frac{1}{f_{1}}$

v' is the object for f₂

u₂ = f₁ – d

$\frac{1}{v} - \frac{1}{u_{2}} = \frac{1}{f_{2}}$

$\frac{1}{v} - \frac{1}{f_{1} - d} = \frac{1}{f_{2}}$

$\frac{1}{v} = \frac{1}{(f_{1} - d)} + \frac{1}{f_{2}}$

$v = \frac{f_{2} (f_{1} - d)}{f_{2} + (f_{1} - d)}$

from the first lens the final image is at a distance

$\frac{f_{1} f_{2} + d (f_{1} - d)}{f_{2} + f_{1} - d} = \frac{{df}_{2} + d (f_{1} - d) + f_{2} f_{1} - f_{2} d}{f_{2} + f_{1} - d}$