
Find the radii of curvature of a convexo-concave converging lens made of glass with refractive index \[n = 1.5\] having focal length of \[24cm\]. One of the radii of curvature is double, then the other?
Answer
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Hint: This formula is used for only the thin lenses. It relates between the focal length, the radii of curvature, and the refractive index of the material of the lens.
Complete step by step answer:
Given, a convexo-concave lens of refractive index, \[n = 1.5\]
Focal length\[f = 24cm\].
Let ${R_1}$ be the radius of curvature of one lens and ${R_2}$ be the radius of curvature of another lens. And it is given that; one of the curvatures is doubled the other then,
$ \Rightarrow {R_2} = 2{R_1}$
We know that for a convexo-concave lens ${R_1}$ is positive and also ${R_2}$ is positive (from Cartesian sign convention)
According to the lens maker’s formula,
$ \Rightarrow \dfrac{1}{f} = \left( {n - 1} \right)\left[ {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right]$
Where $f$is the focal length, $n$is the refractive index, ${R_1},{R_2}$is the radii of curvature.
We can substitute the given values we get,
$\Rightarrow \dfrac{1}{{24}} = \left( {1.5 - 1} \right)\left[ {\dfrac{1}{{{R_1}}} - \dfrac{1}{{2{R_1}}}} \right]$
Let us simplify the given equation,
$ \Rightarrow \dfrac{1}{{24}} = \left( {0.5} \right)\left[ {\dfrac{{2 - 1}}{{2{R_1}}}} \right]$
With the help of the multiplication we get
$ \Rightarrow \dfrac{1}{{24}} = \left[ {\dfrac{{0.5}}{{2{R_1}}}} \right]$
After solving the equation, we get
$ \Rightarrow 2{R_1} = 12$
By substituting and dividing the values we get,
$\therefore {R_1} = 6cm$
And we have, ${R_2} = 2{R_1}$
By substituting and multiplying the values we get,
$ \Rightarrow {R_2} = 2 \times 6$
$\therefore 12cm$
Thus, the radii of curvature of a convexo-concave converging lens are 6cm and 12cm.
Additional information:
A lens is an optical medium bounded by two surfaces of which at least one surface is spherical. lenses commonly used either two spherical surfaces or one spherical surface and one plane surface. These lenses are called spherical lenses. there are different types of lenses, namely biconvex lens, a plano-convex lens, biconcave lens, plano-concave lens, etc.
Note:
- All distances are measured from the pole of the mirror and along the principal axis.
- The distances measured in the direction of the incident light are taken as positive while those measured in the direction opposite to the incident light are taken as negative.
- Heights measured upward and perpendicular to the principal axis are considered positive while those measured downwards are considered negative.
- Manufacturers use the lens maker's formula to design the lenses of required focal lengths from a glass of a given refractive index.
- The refractive index of a medium is a measure of the velocity of light in the given medium.
Complete step by step answer:
Given, a convexo-concave lens of refractive index, \[n = 1.5\]
Focal length\[f = 24cm\].
Let ${R_1}$ be the radius of curvature of one lens and ${R_2}$ be the radius of curvature of another lens. And it is given that; one of the curvatures is doubled the other then,
$ \Rightarrow {R_2} = 2{R_1}$
We know that for a convexo-concave lens ${R_1}$ is positive and also ${R_2}$ is positive (from Cartesian sign convention)
According to the lens maker’s formula,
$ \Rightarrow \dfrac{1}{f} = \left( {n - 1} \right)\left[ {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right]$
Where $f$is the focal length, $n$is the refractive index, ${R_1},{R_2}$is the radii of curvature.
We can substitute the given values we get,
$\Rightarrow \dfrac{1}{{24}} = \left( {1.5 - 1} \right)\left[ {\dfrac{1}{{{R_1}}} - \dfrac{1}{{2{R_1}}}} \right]$
Let us simplify the given equation,
$ \Rightarrow \dfrac{1}{{24}} = \left( {0.5} \right)\left[ {\dfrac{{2 - 1}}{{2{R_1}}}} \right]$
With the help of the multiplication we get
$ \Rightarrow \dfrac{1}{{24}} = \left[ {\dfrac{{0.5}}{{2{R_1}}}} \right]$
After solving the equation, we get
$ \Rightarrow 2{R_1} = 12$
By substituting and dividing the values we get,
$\therefore {R_1} = 6cm$
And we have, ${R_2} = 2{R_1}$
By substituting and multiplying the values we get,
$ \Rightarrow {R_2} = 2 \times 6$
$\therefore 12cm$
Thus, the radii of curvature of a convexo-concave converging lens are 6cm and 12cm.
Additional information:
A lens is an optical medium bounded by two surfaces of which at least one surface is spherical. lenses commonly used either two spherical surfaces or one spherical surface and one plane surface. These lenses are called spherical lenses. there are different types of lenses, namely biconvex lens, a plano-convex lens, biconcave lens, plano-concave lens, etc.
Note:
- All distances are measured from the pole of the mirror and along the principal axis.
- The distances measured in the direction of the incident light are taken as positive while those measured in the direction opposite to the incident light are taken as negative.
- Heights measured upward and perpendicular to the principal axis are considered positive while those measured downwards are considered negative.
- Manufacturers use the lens maker's formula to design the lenses of required focal lengths from a glass of a given refractive index.
- The refractive index of a medium is a measure of the velocity of light in the given medium.
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