Q 9.21 (a)   Determine the ‘effective focal length’ of the combination of the two lenses in Exercise 9.10, if they are placed 8.0cm apart with their principal axes coincident. Does the answer depend on which side of the combination a beam of parallel light is incident? Is the notion of effective focal length of this system useful at all?

Answers (1)
P Pankaj Sanodiya

Here there are two cases, first one is the one when we see it from convex side i.e. Light are coming form infinite and going into convex lens first and then goes to concave lens afterwords. The second case is a just reverse of the first case i.e. light rays are going in concave first.

1)When light is incident on convex lens first

\frac{1}{f}=\frac{1}{v}-\frac{1}{u}

\frac{1}{v}=\frac{1}{f}+\frac{1}{u}

\frac{1}{v}=\frac{1}{30}+\frac{1}{infinite}

v= 30cm

Now this will act as an object for the concave lens.

\frac{1}{f_{concave}}=\frac{1}{v_{fromconcave}}-\frac{1}{u_{fromconcave}}

u_{fromconcave}= 30 - 8 = 22 cm

 \frac{1}{-20}=\frac{1}{v_{fromconcave}}-\frac{1}{22}

\frac{1}{v}=-\frac{1}{220}

v = -220cm

Hence parallel beam of rays will diverge from this point which is (220 - 4 = 216) cm away from the center of the two lenses.

2) When rays fall on concave lens first

\frac{1}{f}=\frac{1}{v}-\frac{1}{u}

\frac{1}{v}=\frac{1}{f}+\frac{1}{u}

\frac{1}{v}=\frac{1}{-20}+\frac{1}{infinite}

v=-20cm

Now this will act as an object for convex lens.

\frac{1}{f_{convex}}=\frac{1}{v_{fromconvex}}-\frac{1}{u_{fromconvex}}

u_{fromconvex}= -20-8=-28cm

\frac{1}{-30}=\frac{1}{v_{fromconvex}}-\frac{1}{-28}

\frac{1}{v_{fromconvex}}=\frac{1}{30}-\frac{1}{28}=-\frac{1}{420}

v_{fromconvex}= -420cm

Hence parallel beam will diverge from this point which is (420 - 4 = 146 cm ) away from the centre of two lenses.

As we have seen for both cases we have different answers so Yes, answer depend on the side of incidence when we talk about combining lenses. i .e. we can not use the effective focal length concept here.

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