Question

Find the focal length of the glass lens in the figure (Figure 1) . Take R₁ = 20 cm and R₂ = 49 cm.

and what is the unit?

Answer 1

Concepts and reason

The concept required to solve this problem is the lens maker formula of biconvex lens.

Initially, write an expression for lens maker formula of biconvex lens. Finally, calculate the focal length of the glass lens.

Fundamentals

The expression for the focal length of the two different radii of curvatures is as follows:

$\frac{1}{f} = \left( {\frac{{{n_{{\rm{glass}}}}}}{{{n_{{\rm{air}}}}}} - 1} \right)\left[ {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}}} \right]$

Here, ${n_{{\rm{glass}}}}$ is the refractive index of glass, ${n_{{\rm{air}}}}$ is the refractive index of air, ${R_1}$ and ${R_2}$ are the radii of curvature, f is the focal length.

Substitute 1.5 for ${n_{{\rm{glass}}}}$ and 1.0 for ${n_{{\rm{air}}}}$ in the equation $\frac{1}{f} = \left( {\frac{{{n_{{\rm{glass}}}}}}{{{n_{{\rm{air}}}}}} - 1} \right)\left[ {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}}} \right]$ .

$\begin{array}{c}\\\frac{1}{f} = \left( {\frac{{1.5}}{1} - 1} \right)\left[ {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}}} \right]\\\\ = \frac{1}{2}\left[ {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}}} \right]\\\end{array}$

Thus, the expression for the reciprocal of focal length is,

$\frac{1}{f} = \frac{1}{2}\left[ {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}}} \right]$

Substitute $20{\rm{ cm}}$ for ${R_1}$ and $- 49{\rm{ cm}}$ for ${R_2}$ in the equation $\frac{1}{f} = \frac{1}{2}\left[ {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}}} \right]$ .

$\begin{array}{c}\\\frac{1}{f} = \frac{1}{2}\left[ {\frac{1}{{20{\rm{ cm}}}} - \frac{1}{{\left( { - 49{\rm{ cm}}} \right)}}} \right]\\\\ = \frac{1}{2}\left[ {\frac{1}{{20{\rm{ cm}}\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1.0{\rm{ cm}}}}} \right)}} + \frac{1}{{49{\rm{ cm}}\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1.0{\rm{ cm}}}}} \right)}}} \right]\\\\ = 3.52{\rm{ }}{{\rm{m}}^{ - 1}}\\\end{array}$

‎Solve for f.

$\begin{array}{c}\\f = \frac{1}{{3.52{\rm{ }}{{\rm{m}}^{ - 1}}}}\\\\ = 0.284{\rm{ m}}\\\\{\rm{ = }}0.284{\rm{ m}}\left( {\frac{{{{10}^2}{\rm{cm}}}}{{1.0{\rm{ m}}}}} \right)\\\\ = 28.40{\rm{ cm}}\\\end{array}$

Ans:

The focal length of the biconvex lens is 28.40 cm.