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Combinations of Lenses & Mirrors

Two identical...

Question

Two identical plano-convex lenses $L_{1} (μ_{1} = 1.4)$ and $L_{2} (μ_{2} = 1.5)$ of radii of curvature $R = 20 c m$ are placed as shown in the figure.
Now, the second lens is shifted vertically downwards by a small distance $4.5 m m$ and the extended parts of $L_{1}$ and $L_{2}$ are blackened as shown in figure. Find the new position of the image of the parallel beam.

200 / 9 c m

behind the lens

2.5 m m

below the principal axis of

L_{1}

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100 / 9 c m

behind the lens

2 m m

below the principal axis of

L_{1}

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200 / 9 c m

in front of the lens

2.5 m m

below the principal axis of

L_{1}

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100 / 9 c m

in front of the lens

2 m m

below the principal axis of

L_{1}

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Solution

The correct option is B $200 / 9 c m$ behind the lens $2.5 m m$ below the principal axis of $L_{1}$
Consider refraction of light from first surface of $L_{1}$ ,

$\frac{1.4}{v_{1}} - \frac{1}{- \infty} = \frac{1.4 - 1}{20}$

Consider refraction of light from second surface of $L_{1}$ ,

$\frac{1}{v_{2}} - \frac{1.4}{v_{1}} = \frac{1 - 1.4}{- \infty}$

Hence, $v_{2} = 50 c m$

Now this image is obtained after refraction from lens $L_{1}$ and lies on the principal axis of $L_{1}$ .

This image gets refracted from $L_{2}$ for which it is at a height of
$4.5 m m$ .

Consider refraction of light from first surface of $L_{2}$ ,

$\frac{1.5}{v_{3}} - \frac{1}{v_{2}} = \frac{1.5 - 1}{- \infty}$

Consider refraction of light from second surface of $L_{2}$ ,

$\frac{1}{v_{4}} - \frac{1.5}{v_{3}} = \frac{1 - 1.5}{- 20}$

$⟹ v_{4} = \frac{200}{9} c m$

Height of final image $= M \times 4.5 m m = 2 m m$

This is the height of object above principal axis of $L_{2}$ .

Hence the final image is $4.5 m m - 2 m m = 2.5 m m$ below principle axis of
$L_{1}$ .

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