Two thin convex lenses of focal lengths f1 and f2 are separated by a horizontal distance d (where d

Question

Two thin convex lenses of focal lengths f1 and f2 are separated by a horizontal distance d (where d < f1, d < f2) and their centres are displaced by a vertical separation Δ as shown in the figure. Taking the origin of coordinates, O, at the centre of the first lens, the x and y-coordinatcs of the focal point of this lens system, for a parallel beam of rays coming from the left, are given by (A) x=(f1 f2/f1 + f2),y =Δ (B) x=(f1 (f2+d)/f1 + f2-d),y =(Δ/f1 + f2) (C) x=(f1 f2+d(f1-d)/f1 + f2-d),y =(Δ (f1 - d)/f1 + f2-d) (D) x=(f1 f2+d(f1-d)/f1 + f2-d), y=0

Tardigrade Admin · Accepted Answer

From the first lens parallel beam of light is focused at its
focus i.e. at a distance

f_{1}

from it. This image

I_{1}

acts as virtual
object for second lens

L_{2}

. Therefore, for

L_{2}

,

u = + (f_{1} - d), f = + f_{2}

∴ \frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{f _{2}} + \frac{1}{f _{1} - d}

Hence,

v = \frac{f _{2} ( f _{1} - d )}{f _{2} + f _{1} - d}

Therefore, x-coordinate of its focal point will be

x = d + v = d + \frac{f _{2} ( f _{1} - d )}{f _{2} + f _{1} - d}

= \frac{f _{1} f _{2} + d ( f _{1} - d )}{f _{1} + f _{2} - d}

Linear magnification for

L_{2}

m = \frac{v}{u} = \frac{f _{2} ( f _{1} - d )}{f _{2} + f _{1} - d} . \frac{1}{f _{1} - d} = \frac{f _{2}}{f _{2} + f _{1} - d}

Therefore, second image will be formed at a distance of m

Δ

or

(\frac{f _{2}}{f _{2} + f _{1} - d}) .Δ

below its optic axis.
Therefore, y-coordinate of the focus of system will be

y = Δ - (\frac{f _{2} Δ}{f _{2} + f _{1} - d})

or

y = \frac{( f _{1} - d ) .Δ}{f _{2} + f _{1} - d}

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

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

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

$x = \frac{f _{1} f _{2} + d ( f _{1} - d )}{f _{1} + f _{2} - d}, y = 0$

x=f1​+f2​f1​f2​​,y=Δ

x=f1​+f2​−df1​(f2​+d)​,y=f1​+f2​Δ​

x=f1​+f2​−df1​f2​+d(f1​−d)​,y=f1​+f2​−dΔ(f1​−d)​

x=f1​+f2​−df1​f2​+d(f1​−d)​,y=0

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

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

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

$x = \frac{f _{1} f _{2} + d ( f _{1} - d )}{f _{1} + f _{2} - d}, y = 0$