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Worked Examples · Example 9.8

Q.Find the position of the image formed by the lens combination given in the Fig. 9.20.

Figure 9.20
Figure 9.20
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Trace the light through the three lenses in turn, each image becoming the next object: v1=+15v_1=+15 cm, then v2=∞v_2=\infty (a parallel beam), then v3=+30v_3=+30 cm. The final image is real, 30 cm to the right of the third lens.

Concept understanding

For thin lenses in a row on one axis, refraction is handled one lens at a time with

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

using the image of one lens as the object for the next and accounting for the separation between lenses. Sign convention: distances measured along the incident light (left→\toright) are positive; convex f>0f>0, concave f<0f<0.

Step-by-step (setup: object 30 cm left of lens 1; lenses at 0, 5 cm, 10 cm with f1=+10f_1=+10, f2=−10f_2=-10, f3=+30f_3=+30)

Lens 1 (convex, f1=+10f_1=+10 cm): u1=−30u_1=-30 cm.

1v1=1f1+1u1=110−130=230  ⇒  v1=+15 cm.\frac{1}{v_1}=\frac{1}{f_1}+\frac{1}{u_1}=\frac{1}{10}-\frac{1}{30}=\frac{2}{30}\;\Rightarrow\; v_1=+15\text{ cm}.

A real image forms 15 cm to the right of lens 1.

Lens 2 (concave, f2=−10f_2=-10 cm): lens 2 is 5 cm right of lens 1, so this image is 15−5=1015-5=10 cm to the right of lens 2 — a virtual object, u2=+10u_2=+10 cm.

1v2=1f2+1u2=−110+110=0  ⇒  v2=∞.\frac{1}{v_2}=\frac{1}{f_2}+\frac{1}{u_2}=-\frac{1}{10}+\frac{1}{10}=0\;\Rightarrow\; v_2=\infty. …

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