91Ó°ÊÓ

• Magnitude of the radius of curvature $R=10\text{cm}$,
• Figure 34-30 showing four lenses are given.

The focal length of each lens by using the lens maker formula is calculated. Then rank them according to their magnitudes.

Formula:

The lens maker equation,$\frac{1}{f}=\left(\mathrm{μ}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)······\left(1\right)$

Here, $\mathrm{μ}=\text{refractiveindexoflens}$

$R_1=\text{radiusofcurvatureofincidentlight}$

$R_2=\text{radiusofcurvaturethroughwhichlightgoesout}$

Refractive index of glass,${\mathrm{μ}}_{glass}=1.5$

Case 1:

$R_1=10\text{cm};R_2=∞$

The focal length for the above data can be given using equation (1) as follows:

$\begin{array}{rcl}\frac{1}{f_a}& =& \left(1.5-1\right)\left(\frac{1}{10}-\frac{1}{∞}\right)\\ f_a& =& 20\text{cm}\end{array}$

Case 2:

$R_1=∞;R_2=-10\text{cm}$

The focal length for the above data can be given using equation (1) as follows:

$\begin{array}{rcl}\frac{1}{f_b}& =& \left(1.5-1\right)\left(\frac{1}{∞}-\frac{1}{-10}\right)\\ f_b& =& -20\text{cm}\\ & & \end{array}$

Case 3:

$R_1=10\text{cm};R_2=-10\text{cm}$

The focal length for the above data can be given using equation (1) as follows:

$\begin{array}{rcl}\frac{1}{f_d}& =& \left(1.5-1\right)\left(\frac{1}{10}-\frac{1}{-10}\right)\\ f_d& =& 10\text{cm}\\ & & \end{array}$

Here,$R_2<R_1$

Hence, the focal length$f_d$will be greater than the focal lengths$f_a,f_b\text{and}{f}_c$.

Therefore, the rank of the lenses according to the magnitude of the focal length is $f_d>f_a=f_b>f_c$.