The problem statement:

Two convex lenses, each of focal length 10cm, are placed at a separation of 15cm with their principal axes coinciding.

  1. Show that a light beam coming parallel to the principal axis diverges as it comes out of the lens system.

My approach:

I've applied the focal length equivalent formula, $$\frac{1}{f_{eq}}=\frac{1}{f_{1}}+\frac{1}{f_{2}}-\frac{d}{f_{1}*f_{2}}$$ $$f_{eq}=+20 cm$$

Upon looking at the focal length I'm confused as my answer's focal length is positive(converging) but the nature of the given lens is diverging in the first subpart.

Is my approch right or I'm using wrong sign convention?

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    $\begingroup$ Forget about the formula and draw a ray diagram to show the position of the intermediate image produced by the first lens which then acts as the object for the second lens. $\endgroup$ – Farcher Feb 28 '19 at 16:20
  • $\begingroup$ I've done that but is the formula incorrect because I can't use the whole process for large problems because its time consuming. $\endgroup$ – Sahil Silare Feb 28 '19 at 17:17
  • $\begingroup$ Interesting configuration. After the focus of positive lens the rays are divergent. For this configuration the focus or more precisely the back focal plane is inside between the two lenses meaning after the second lens rays are divergent. Just calculate the back principle plane position and the position of the back focal plane. Your formula gives effective focusing length. $\endgroup$ – user591849 Mar 1 '19 at 4:36
  • $\begingroup$ @user591849 Can you explain what's back focal plane? $\endgroup$ – Sahil Silare Mar 1 '19 at 6:20
  • $\begingroup$ A system of lenses can be combined into an effective system of single lens. When we combine the two lenses into one there will be just two focal planes and two principle planes. The back focal plane is the plane after the lens. The back focal plane of the combined system happens to be before the physical location of the second lens. $\endgroup$ – user591849 Mar 1 '19 at 8:08

I believe your calculations are correct you just didn't complete it. The formula you used are to find a lens whose focal length would produce an image at the same position when replacing the two lenses, but where is it going to be placed ?!

The formula that is used to find at what distance should this equivalent lens be placed from the second lens is $$ x = \dfrac{d ~ f_{eq}}{f_1}$$

It is better to use the lens maker's equation twice, which is given by $$ \dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}$$ Where, $u$ is the distance between the object and the lens, and $v$ is the distance between the image and the lens, and if you showed that the final image is formed in front of the second lens then you would have proved that this system is actually a diverging system "for objects that is infinitely distant from the second lens".

Knowing that the light incident on the first lens is parallel to the principal-axis, this means that the object is infinitely distant from the lens, and substituting in the lens maker's equation, it is obvious that the image would be formed at the focal point of the first lens ``the image would be real and formed behind the lens''.

Thus, the image would be formed in front the focal point of the next lens, which is separated a distance 15 cm from the first lens, which means that the image which now serves as a real object for the second lens, is at a distance of 5 cm from the second lens.

A convex lens, is capable of producing both real and virtual images "it can either converge or diverge the light incident on it", depending on the position of the object with the respect to its focal point.

And since the image is in front the focal point of the second lens, then the second lens would diverge the image formed by the first lens "which is the object for the second lens", and if you want to find where the image is formed, then you can use the lens maker's equation once again, substituting the focal length value and the distance $u$ to be 5 cm, then we can find $v$ which would have a negative sign of 10 cm.

However, the final image of this optical system is a diverging, where the final image is formed in front the second lens.

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    $\begingroup$ I can obviously use Lens Maker's formula twice but I find it time taking and I can't use it for 3-4 lenses combination. Hence, I was looking for a shorter and faster approach. Thanks! $\endgroup$ – Sahil Silare Mar 2 '19 at 11:03
  • $\begingroup$ I believe you can judge by knowing where the image would be with the respect to the focal point of a convex lens, and a diverging lens would always produced a virtual image, for example in this case, as you know that the image by the first lens is at the focal point of the first lens, which is in front the focal point of the other lens, thus the final image is divergent. $\endgroup$ – A Onsi Mar 2 '19 at 14:53

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