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Question

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A. 9cm and 18cm

B. 6cm and 12cm

C. 3cm and 6cm

D. 4.5 and 9cm

Answer
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$\dfrac{1}{f}=(\mu -1)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)$

In the question, it is given that the ratio of radii of a converging lens is 1:2

The focal length of the lens is 6cm

And the refractive index is 1.5.

Let the radius ${{R}_{1}}=R$

Then${{R}_{2}}$, must be negative, and two times of ${{R}_{1}}$, as given in the ratio,

${{R}_{2}}=-2R$.

Now applying Lens maker formula,

The formula is,

$\dfrac{1}{f}=(\mu -1)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)$

$\dfrac{1}{6}=(1.5-1)\left( \dfrac{1}{R}+\dfrac{1}{2R} \right)$, (on replacing with the values given and considered)

$\Rightarrow \dfrac{1}{6}=(0.5)\left( \dfrac{3}{2R} \right)$,

$\Rightarrow \dfrac{1}{6}=\left( \dfrac{1.5}{2R} \right)$,

$\Rightarrow R=4.5cm$

${{R}_{1}}=4.5cm$.

${{R}_{2}}=9cm$.

When any ray of light travels parallel to a lens principal axis, and the lens converges those rays of light to its middle, that type of lens is called a converging lens. These lenses are thinner at the periphery and thicker in the middle.

Lens maker formula is a formula that gives us the relationship between focal length in terms of the refractive index and radius of the two curvatures of the lens.

In the formula, $\dfrac{1}{f}=(\mu -1)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)$, ‘f’ is the focal length, ‘$\mu $’, is the refractive index and ${{R}_{1}}$ is the radius of curvature of one side of the lens and ${{R}_{1}}$ is radius of curvature of the second side of lens.