Question

Find the focal length of the glass lens in the figure (Figure 1). Take R1...

Find the focal length of the glass lens in the fig

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?

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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:

1f=(nglassnair1)[1R11R2]\frac{1}{f} = \left( {\frac{{{n_{{\rm{glass}}}}}}{{{n_{{\rm{air}}}}}} - 1} \right)\left[ {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}}} \right]

Here, nglass{n_{{\rm{glass}}}} is the refractive index of glass, nair{n_{{\rm{air}}}} is the refractive index of air, R1{R_1} and R2{R_2} are the radii of curvature, f is the focal length.

Substitute 1.5 for nglass{n_{{\rm{glass}}}} and 1.0 for nair{n_{{\rm{air}}}} in the equation 1f=(nglassnair1)[1R11R2]\frac{1}{f} = \left( {\frac{{{n_{{\rm{glass}}}}}}{{{n_{{\rm{air}}}}}} - 1} \right)\left[ {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}}} \right] .

1f=(1.511)[1R11R2]=12[1R11R2]\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,

1f=12[1R11R2]\frac{1}{f} = \frac{1}{2}\left[ {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}}} \right]

Substitute 20cm20{\rm{ cm}} for R1{R_1} and 49cm - 49{\rm{ cm}} for R2{R_2} in the equation 1f=12[1R11R2]\frac{1}{f} = \frac{1}{2}\left[ {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}}} \right] .

1f=12[120cm1(49cm)]=12[120cm(102m1.0cm)+149cm(102m1.0cm)]=3.52m1\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.

f=13.52m1=0.284m=0.284m(102cm1.0m)=28.40cm\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.

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