Optics problem!

$With\quad the\quad definition\quad of\quad focus,\quad we\quad take\quad a\quad parallel\quad beam\\ of\quad light\quad at\quad the\quad first\quad lens\quad of\quad f=\quad 10cm.\\ \\ By\quad definition\quad the\quad rays\quad will\quad try\quad to\quad intersect\quad at\quad 10cm.\\ \\ But\quad \because \quad 10cm>\quad d\quad =\quad 5cm\quad where\quad d\quad is\quad the\quad seperation.\\ Thus\quad we\quad have\quad a\quad virtual\quad object\quad at\quad +5cm\quad for\quad the\quad \\ second\quad lens\quad of\quad f=\quad -10cm.\\ \\ \frac { 1 }{ v } \quad -\quad \frac { 1 }{ 5 } \quad =\quad \frac { 1 }{ -10 } \\ v\quad =\quad 10cm.\\ \\ Now\quad repeating\quad the\quad same\quad for\quad third\quad lens\quad of\quad \\ f=\quad 10cm,\quad we\quad have\quad u\quad =\quad 10\quad -\quad 5\quad =\quad +5cm.\\ \\ \frac { 1 }{ v } -\frac { 1 }{ 5 } =\frac { 1 }{ 10 } \\ v\quad =\quad \frac { 10 }{ 3 } cm\\ v=\quad \boxed { 3.3333cm }$

We are given powers as 10D, -10D, 10D.

so, f=10cm, -10cm,10cm

Let a parallel beam of light strike lens 1. The point at which this beam converges at the principal axis is the secondary focus of the whole system, which we require.

In $1^{st}$ refraction, ray tries to converge at 10cm from lens, but we have lens 2 at 5cm. So we have a virtual object for lens 5cm in front of it.

1/v - 1/5 = -1/10

v=10cm

But we have another lens at 5cm. So we have a virtual object for lens 3, 5cm ahead.

1/v - 1/5 = 1/10

v=10/3 = 3.33 :)