Collection of Solved Problems in Physics

A Plano-concave Lens

Task number: 1964

A plano-concave lens made of glass of refractive index of \(1{.}5\) is located in the air. Its focal length is \(−10 \mathrm{cm}\). Determine its radius of curvature.

• Notation

  \(n_{g}=1{.}5\)	refractive index of glass
  \(f=10 \mathrm{cm}\)	focal length
  \(r= ?\)	radius of curvature

• Hint 1

  Each lens has two radii of curvature. Why are we looking only for the value of one of them in this case?

• Solution of Hint 1

  We already know the value of the radius of curvature \(r_{1}\) of the plane (first) surface because it is approaching infinity. That is why we are only looking for the radius of curvature of the concave (second) surface \(r_{2}=r\).

• Hint 2

  Let us assume that it is a thin lens. The radii of curvature affect the focal length of the lens. Could you describe this fact mathematically?

• Solution of Hint 2

  Change in the focal length of the lens \(f\) in dependence on the changes in the radii of curvature of optical surfaces \(r_{1}\) and \(r_{2}\) is defined by the lensmaker’s equation. Let’s write it for a thin lens. \[\frac{1}{f}=\left(\frac{n_{g}}{n_{a}}−1\right)\left(\frac{1}{r_{1}}+\frac{1}{r_{2}}\right),\] where \(n_{g}\) is the refractive index of glass and \(n_{a}\) is the refractive index of air.

• Hint 3

  How should we adjust the lensmaker’s equation, so that the focal length depends on only the one radius of curvature \(r_{2}=r\)? Then express this radius and calculate it.

• Solution of Hint 3

  We write \(r\) instead of \(r_{2}\) in the lensmaker’s equation: \[\frac{1}{f}=\left(\frac{n_{g}}{n_{a}}−1\right)\left(\frac{1}{r_{1}}+\frac{1}{r_{2}}\right),\] \[\frac{1}{f}=\left(\frac{n_{g}}{n_{a}}−1\right)\left(\frac{1}{r_{1}}+\frac{1}{r}\right).\]

  Radius \(r_{1}\) is approaching infinity, which means that the fraction \(\frac{1}{r_{1}}\) is approaching zero. Thus, we can write:

  \[\frac{1}{f}=\left(\frac{n_{g}}{n_{a}}−1\right)\left(0+\frac{1}{r}\right),\] \[\frac{1}{f}=\left(\frac{n_{g}}{n_{a}}−1\right)\frac{1}{r}.\]

  We multiply both sides of the equation by \(r\cdot f\): \[r=\left(\frac{n_{g}}{n_{a}}−1\right)f.\]

  Numerical solution

  We know from the task assignment:

  \(n_{g}=1{.}5\) (refractive index of glass)

  \(f=−10 \mathrm{cm}\) (focal length of the lens)

  We look up the value of the refractive index of air in the Tables:

  \(n_{a}=1{.}0\)

  We substitute for the values:

  \(r=\left(\frac{n_{g}}{n_{a}}−1\right)f=\left(\frac{1{.}5}{1{.}0}−1\right)\cdot \left(−10\right) \mathrm{cm}=0{.}5\cdot \left(−10\right) \mathrm{cm}=−5 \mathrm{cm}.\)

• Note

  The value of the radius of curvature is negative because it is the radius of curvature of the concave surface (see the sign convention mentioned in Imaging a Bat by a Spherical Mirror and a Converging Lens, Radius of curvature and focal length).

• Overall Solution

  The lensmaker’s equation for a thin lens

  Change in the focal length of the lens \(f\) in dependence on the changes in the radii of curvature of optical surface \(r_{1}\) and \(r_{2}\) is defined by the lensmaker’s equation. Let’s assume that we have a thin lens: \[\frac{1}{f}=\left(\frac{n_{g}}{n_{a}}−1\right)\left(\frac{1}{r_{1}}+\frac{1}{r_{2}}\right),\] where \(n_{g}\) is the refractive index of glass and \(n_{a}\) is the refractive index of air.

   

  Expressing the radius of curvature

  Radius of curvature \(r_{1}\) is approaching infinity, which means that fraction \(\frac{1}{r_{1}}\) is approaching zero. Therefore, this fraction will „disappear“ from the equation:

  \[\frac{1}{f}=\left(\frac{n_{g}}{n_{a}}−1\right)\frac{1}{r_{2}}.\]

  Denote radius of curvature \(r_{2}\) by \(r\) (\(r_{2}=r\)):

  \[\frac{1}{f}=\left(\frac{n_{g}}{n_{a}}−1\right)\frac{1}{r}.\]

  Multiply both sides of the equation by \(r\cdot f\): \[r=\left(\frac{n_{g}}{n_{a}}−1\right)f.\]

  Numerical solution

  We know from the task assignment:

  \(n_{g}=1{.}5\) (refractive index of glass)

  \(f=−10 \mathrm{cm}\) (focal length of the lens)

  We look up the value of the refractive index of air in the Tables:

  \(n_{a}=1{.}0\)

  We substitute for the values:

  \(r=\left(\frac{n_{g}}{n_{a}}−1\right)f=\left(\frac{1{.}5}{1{.}0}−1\right)\cdot \left(−10\right) \mathrm{cm}=0{.}5\cdot \left(−10\right) \mathrm{cm}=−5 \mathrm{cm}.\)

  It is the radius of curvature of the concave surface, therefore, its value is negative.

• Answer

  The value of the radius of curvature of the plano-concave lens is \(−5 \mathrm{cm}\).