Figure shows a biconvex lens of having radius of curvatures as R1 and R2 , in the ray depicted shouldnt the meeting point of the rays be at 2F and the object should be placed at -2F , where F is the focal length from the lens maker formula for thin lenses ? The figure depicted -2F1 going to 2F1 this is wrong isnt ? It should be F in place of F1 ,where 1/F= 1/F1 +1/F2 isnt ? And is the F1 and F2 called the focal point of lenses having one radius of curvature as infinity ?is that the definition of first and secondary focus(other than the parallel rays meeting one)? 
Is this given ray diagram for a biconvex lens correct
geometric-opticslenses
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All you need to know about the chief ray is that it will show up in the right place at the focal plane. Any path that gets there is valid. You will see that this is indeed the case for your example - the ray starts up at the top of the image (tip of the arrow) and passes through the tip of the arrow in the "internal image" plane.
Any pair of straight lines that gets from A to B is valid - although when you are trying to construct where the internal focal plane is, the construction as drawn is not useful, once you know where it is you can draw any of a family of lines, including the one shown.
As for the second half of the trace: if the chief ray passes the aperture stop on the optical axis, then it must also pass the "image" of the aperture stop at that place - which gives rise to the trace that gets you the rest of the image.
So - it follows the normal rules; but it is unorthodox...
The lens equation and the ray tracing does agree. Let's denote the distance from the object and image distances to the first convex lens as $o_1$ and $i_1$ respectively. Let's denote the object and image distances to the second concave lens as $o_2$ and $i_2$. The focal lengths are $f_1=2 \text{ cm}$ and $f_2=-2 \text{ cm}$.
The lens equation reads $\frac{1}{o}+\frac{1}{i}=\frac{1}{f}$.
For the first lens, $\frac{1}{3\text{ cm}}+\frac{1}{i_1}=\frac{1}{2\text{ cm}}$.
Solving for the image distance yields $i_1=6\text{ cm}$.
The image of the first lens becomes the object for the second lens, and since the two lenses are $4\text{ cm}$ apart, we have $o_2 = 4\text{ cm}-i_1 = -2\text{ cm}$.
The lens equation for the second lens reads as $\frac{1}{-2\text{ cm}}+\frac{1}{i_2}=\frac{1}{-2\text{ cm}}$.
Thus $1/i_2 = 0$, and $i_2=\infty$. There is neither a real image (converging rays in front of the second lens) nor a virtual image that is formed (diverging rays). Rather the rays are parallel.
Well, we can see this agrees with the ray tracing diagram (made with OpticalRayTracer software ver. 9.6):
Best Answer
F$_1$ and F$_2$ are the principal foci. They are the same distance, f, (the focal length) either side of the optical centre of the lens.
I can therefore make no sense of your equation "1/F= 1/F1 +1/F2".
The rays on the diagram are roughly correct. An object in a plane at distance 2f from the optical centre of the lens should form an image on the plane at a distance 2f from the optical centre but to the other side of the lens. You can show this by putting $u=2f$ into the equation $$\frac 1u + \frac 1v = \frac 1f\ \ \ \ \ \ \ \text{[real-is-positive convention]}$$ or, equivalently, by putting $u=-2f$ into the equation $$\frac 1u + \frac 1f = \frac 1v\ \ \ \ \ \ \ \text{[cartesian convention]}$$
I assume that 2F$_1$ and 2F$_2$ are supposed to be points on the lens axis a distance 2f from the optical centre of the lens, on either side of the lens. 2F$_2$ seems to be marked too close to the lens.
When I first looked at the diagram I was misled by the labels C$_1$ and C$_2$, expecting them to stand for the centres of curvature of the two faces of the lens. But these will not be at distance 2f from the centre of the lens, as on the diagram. [It would require a refractive index of 2 for the lens material, as you can show from the lens-makers' formula.] Clearly C$_1$ and C$_2$ have nothing to do with centres of curvature!