Question

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 18cm
B. 6cm and 12cm
C. 3cm and 6cm
D. 4.5 and 9cm

Answer 1

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

Complete step by step answer:
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”.

Additional Information:
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.

Note:
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.