91Ó°ÊÓ

Object’s distance is s and image distance is s’.

The relationship is$\frac{1}{s}+\frac{1}{s\text{'}}=\frac{1}{\mathit{f}}$

$\frac{1}{\mathit{f}}=\left(\mathit{n}-1\right)(\frac{1}{{\mathit{R}}_{\mathbf{1}}}-\frac{1}{{\mathit{R}}_{\mathbf{2}}})$where n is the refraction index and R is the curvatures of the lens.

Using therelationship between object and image distance we get,

$\frac{1}{s_1}=\frac{1}{{\mathit{f}}_1}-\frac{1}{s{\text{'}}_1}$

Now adding these two for both ${\mathit{s}}_{\mathbf{1}}\mathbf{\&}{\mathit{s}}_{\mathbf{2}}$ will result in,

role="math" localid="1663919385591" $$\begin{gathered} \frac{1}{s_1}+\frac{1}{s{\text{'}}_2}=\frac{1}{\mathit{f}} \\ \left(\frac{1}{{\mathit{f}}_1}-\frac{1}{s{\text{'}}_1}\right)+\left(\frac{1}{{\mathit{f}}_2}-\frac{1}{s_2}\right)=\frac{1}{\mathit{f}} \\ \left(\frac{1}{{\mathit{f}}_1}-\frac{1}{{\mathit{f}}_2}\right)+\left(\frac{1}{s{\text{'}}_1}-\frac{1}{s_2}\right)=\frac{1}{\mathit{f}} \end{gathered}$$

Since the two lenses have different focal lengths and are touching each other.Here the distance of image for first lens is same as the distance of object for second lens.Therefore $s{\text{'}}_1=-s_2$

Therefore substituting the value we get,

$$\begin{gathered} \frac{1}{\mathit{f}}=\left(\frac{1}{{\mathit{f}}_2}-\frac{1}{{\mathit{f}}_1}\right)+\left(\frac{1}{s{\text{'}}_1}-\frac{1}{s_2}\right) \\ =\left(\frac{1}{{\mathit{f}}_2}-\frac{1}{{\mathit{f}}_1}\right)+\left(\frac{1}{-s_2}-\frac{1}{s_2}\right) \\ \frac{1}{\mathit{f}}=\frac{1}{{\mathit{f}}_2}+\frac{1}{{\mathit{f}}_1} \end{gathered}$$

Hence proved.

Using the lensmaker’s formula, the focal length of the lenses is derived.

i)The focal length of the glass,

$$\begin{gathered} \frac{1}{{\mathit{f}}_{\mathbf{g}}}=\left(\mathit{n}-1\right)\left(\frac{1}{{\mathit{R}}_{\mathbf{1}}}-\frac{1}{{\mathit{R}}_{\mathbf{2}}}\right) \\ =\left(1.55-1.0\right)\left(\frac{1}{4.5\mathbf{cm}}-\frac{1}{9\mathbf{cm}}\right) \\ =\frac{1}{16.4\mathbf{cm}} \end{gathered}$$

ii)The focal length of the ${\mathbf{CCI}}_{\mathbf{4}}$.

$$\begin{gathered} \frac{1}{{\mathit{f}}_{\mathit{w}}}=\left({\mathit{n}}_{\mathbf{w}}-1\right)\left(\frac{1}{{\mathit{R}}_{\mathbf{1}}}-\frac{1}{{\mathit{R}}_{\mathbf{2}}}\right) \\ =\left(1.46-1.0\right)\left(\frac{1}{9\mathbf{cm}}-\frac{1}{∞}\right) \\ =\frac{1}{19.6\mathbf{cm}} \end{gathered}$$

Therefore the combined focal length will be,

$$\begin{gathered} \frac{1}{\mathit{f}}=\frac{1}{{\mathit{f}}_{\mathbf{g}}}+\frac{1}{{\mathit{f}}_{\mathbf{w}}} \\ \frac{1}{\mathit{f}}=\frac{1}{16.4}+\frac{1}{19.6} \\ \frac{1}{\mathit{f}}=\frac{1}{8.93\mathit{c}\mathit{m}} \\ \mathit{f}=8.93\mathit{c}\mathit{m} \end{gathered}$$