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Geometrical Optics — Physics STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 SciencePhysicsGeometrical Optics4 Marks
Question
Consider the situation described in the previous problem. Where should a point source be placed on the principal axis so that the two images form at the same place?
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Answer

Let the source be placed at a distance ‘x’ from the lens as shown, so that images formed by both coincide.

For the lens, $\frac{1}{\text{v}_\ell}-\frac{1}{-\text{x}}=\frac{1}{15}\Rightarrow\text{v}_\ell=\frac{15\text{x}}{\text{x}-15} \ ...(1)$

For the mirror, $\text{u}=-(50-\text{x}), \ \text{f}=-10\text{cm}$

So, $\frac{1}{\text{v}_\text{m}}+\frac{1}{-(50-\text{x})}=-\frac{1}{10}$

$\Rightarrow\frac{1}{\text{v}_\text{m}}=\frac{1}{-(50-\text{x})}-\frac{1}{10}$

$\text{v}_\text{m}=\frac{10(50-\text{x})}{\text{x}-40} \ ...(2)$

Since the lens and mirror are 50cm apart,

$\text{v}_\ell-\text{v}_{\text{m}}=50\Rightarrow\frac{15\text{x}}{\text{x}-15}-\frac{10(50-\text{x})}{(\text{x}-40)}=50$

$\Rightarrow\text{x}=30\text{cm}.$

So, the source should be placed 30cm from the lens.

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