Determine the ‘effective focal length’ of the combination of the two lenses in if they are placed 8.0 cm 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.
Exercise
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Here, Focal length of first lens, $f_1 = 30$ cm
Focal length of second lens, $f_2 = -20$ cm
Distance between the lens, $d = 8.0$ cm
Let a parallel beam be incident on the convex lens first. If second lens were absent, then
$\text{u}_1=\infty \ \text{and} \ \text{f}_1=30 \ \text{cm}$
In the formula, as
$\frac{1}{\text{v}_1}-\frac{1}{\text{u}_1}=\frac{1}{\text{f}_1}$
$\therefore \ \frac{1}{\text{v}_1}-\frac{1}{\infty}=\frac{1}{30}$
i.e.$, v_1 = 30\ cm$
This image would now act as a virtual object for second lens.
$\therefore \ \text{u}_2=+(30-8)=+22 \ \text{cm}$
$f_2 = -20 cm$
Since,
$\frac{1}{\text{v}_2}=\frac{1}{\text{f}_2}+\frac{1}{\text{u}_2}$
$\frac{1}{\text{v}_2}=\frac{1}{-20}+\frac{1}{22}$
$=\frac{-11+10}{220}=\frac{-1}{220}$
$\therefore$ Parallel incident beam would appear to diverge from a point 220 - 4 = 216 cm from the centre of the two lens system.
Assume that a parallel beam of light from the left is incident first on the concave lens.
$\therefore \ \text{u}_1=-\infty \ \text{and} \ \text{f}_1=-20 \ \text{cm}$
As, $\frac{1}{\text{v}_1}-\frac{1}{\text{u}_1}=\frac{1}{\text{f}_1}$
$\therefore$ $\frac{1}{\text{v}_1}=\frac{1}{\text{f}_1}+\frac{1}{\text{u}_1}$
$=\frac{1}{-20}+\frac{1}{-\infty}$
$=\frac{1}{-20}$
$v_1 = -20$ cm
This image acts as a real object for the second lens
$u_2= -(20 + 8) = -28$ cm and,
$f_2= 30$ cm
Since, $\frac{1}{\text{v}_2}-\frac{1}{\text{u}_2}=\frac{1}{\text{f}_2}$
$\therefore \ \frac{1}{\text{v}_2}=\frac{1}{\text{f}_2}+\frac{1}{\text{u}_2}$
$=\frac{1}{30}-\frac{1}{28}$
$=\frac{14-15}{420}$
$\text{v}_2=-420 \ \text{cm}$
$\therefore$ The parallel beam appears to diverge from a point 420 - 4 = 416 cm, on the left of the centre of the two lens system.
We finally conclude that the answer depends on the side of the lens system where the parallel beam is incident. Therefore, the notion of effective focal length does not seem to be meaningful here.
Focal length of second lens, $f_2 = -20$ cm
Distance between the lens, $d = 8.0$ cm
Let a parallel beam be incident on the convex lens first. If second lens were absent, then
$\text{u}_1=\infty \ \text{and} \ \text{f}_1=30 \ \text{cm}$
In the formula, as
$\frac{1}{\text{v}_1}-\frac{1}{\text{u}_1}=\frac{1}{\text{f}_1}$
$\therefore \ \frac{1}{\text{v}_1}-\frac{1}{\infty}=\frac{1}{30}$
i.e.$, v_1 = 30\ cm$
This image would now act as a virtual object for second lens.
$\therefore \ \text{u}_2=+(30-8)=+22 \ \text{cm}$
$f_2 = -20 cm$
Since,
$\frac{1}{\text{v}_2}=\frac{1}{\text{f}_2}+\frac{1}{\text{u}_2}$
$\frac{1}{\text{v}_2}=\frac{1}{-20}+\frac{1}{22}$
$=\frac{-11+10}{220}=\frac{-1}{220}$
$\therefore$ Parallel incident beam would appear to diverge from a point 220 - 4 = 216 cm from the centre of the two lens system.
Assume that a parallel beam of light from the left is incident first on the concave lens.
$\therefore \ \text{u}_1=-\infty \ \text{and} \ \text{f}_1=-20 \ \text{cm}$
As, $\frac{1}{\text{v}_1}-\frac{1}{\text{u}_1}=\frac{1}{\text{f}_1}$
$\therefore$ $\frac{1}{\text{v}_1}=\frac{1}{\text{f}_1}+\frac{1}{\text{u}_1}$
$=\frac{1}{-20}+\frac{1}{-\infty}$
$=\frac{1}{-20}$
$v_1 = -20$ cm
This image acts as a real object for the second lens
$u_2= -(20 + 8) = -28$ cm and,
$f_2= 30$ cm
Since, $\frac{1}{\text{v}_2}-\frac{1}{\text{u}_2}=\frac{1}{\text{f}_2}$
$\therefore \ \frac{1}{\text{v}_2}=\frac{1}{\text{f}_2}+\frac{1}{\text{u}_2}$
$=\frac{1}{30}-\frac{1}{28}$
$=\frac{14-15}{420}$
$\text{v}_2=-420 \ \text{cm}$
$\therefore$ The parallel beam appears to diverge from a point 420 - 4 = 416 cm, on the left of the centre of the two lens system.
We finally conclude that the answer depends on the side of the lens system where the parallel beam is incident. Therefore, the notion of effective focal length does not seem to be meaningful here.
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