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# Radii of curvature of a converging lens are in the ratio 1:2. Its focal length is 6cm and refractive index is 1.5. Then its radii of curvature are _________ respectively.A. 9cm and 18cmB. 6cm and 12cmC. 3cm and 6cmD. 4.5 and 9cm

Hint: The ratio of radius, focal length, and refractive index is given. From the ratio of radii try to consider the values of the radius. Then apply the lens maker formula. And then replace the variables with the values that are given and the values that we have considered.

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

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)
$\Rightarrow \dfrac{1}{6}=(0.5)\left( \dfrac{3}{2R} \right)$,
$\Rightarrow \dfrac{1}{6}=\left( \dfrac{1.5}{2R} \right)$,
$\Rightarrow R=4.5cm$

${{R}_{1}}=4.5cm$.
${{R}_{2}}=9cm$.
So, the correct answer is “Option D”.

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.