A diverging lens of focal length 20cm and a converging lens of focal length 30cm are placed 15cm apart with their principal axes coinciding. Where should an object be placed on the principal axis so that its image is formed at infinity?
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Given that, $f_1 = $ focal length of converging lens $= 30cm$
$f_2 =$ focal length of diverging lens $= -20cm$
and d = distance between them $= 15cm$
Let, F = equivalent focal length
So, $\therefore \ \frac{1}{\text{F}}=\frac{1}{\text{f}_1}+\frac{1}{\text{f}_2}-\frac{\text{d}}{\text{f}_1\text{f}_2}$
$\Rightarrow\frac{1}{30}+\Big(-\frac{1}{20}\Big)-\Big(\frac{15}{30(-200)}\Big)=\frac{1}{120}$
$\Rightarrow\text{F}=120\text{cm}$
⇒ The equivalent lens is a converging one.
Distance from diverging lens so that emergent beam is parallel (image at infinity),
$\text{d}_1=\frac{\text{dF}}{\text{f}_1}=\frac{15\times120}{30}=60\text{cm}$
It should be placed 60cm left to diverging lens.
⇒ Object should be placed (120 - 60) = 60cm from diverging lens.
Similarly, $\text{d}_2=\frac{\text{dF}}{\text{f}_2}=\frac{15\times120}{20}=90\text{cm}$
So, it should be placed 90cm right to converging lens.
⇒ Object should be placed (120 + 90) = 210cm right to converging lens.
$f_2 =$ focal length of diverging lens $= -20cm$
and d = distance between them $= 15cm$
Let, F = equivalent focal length
So, $\therefore \ \frac{1}{\text{F}}=\frac{1}{\text{f}_1}+\frac{1}{\text{f}_2}-\frac{\text{d}}{\text{f}_1\text{f}_2}$
$\Rightarrow\frac{1}{30}+\Big(-\frac{1}{20}\Big)-\Big(\frac{15}{30(-200)}\Big)=\frac{1}{120}$
$\Rightarrow\text{F}=120\text{cm}$
⇒ The equivalent lens is a converging one.
Distance from diverging lens so that emergent beam is parallel (image at infinity),
$\text{d}_1=\frac{\text{dF}}{\text{f}_1}=\frac{15\times120}{30}=60\text{cm}$
It should be placed 60cm left to diverging lens.
⇒ Object should be placed (120 - 60) = 60cm from diverging lens.
Similarly, $\text{d}_2=\frac{\text{dF}}{\text{f}_2}=\frac{15\times120}{20}=90\text{cm}$
So, it should be placed 90cm right to converging lens.
⇒ Object should be placed (120 + 90) = 210cm right to converging lens.
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