Q90P - University Physics with Modern Physics

Question

(a) Prove that when two thin lenses with focal lengths ${\mathit{f}}_{\mathbf{1}}$ and ${\mathit{f}}_{\mathbf{2}}$ are placed in contact, the focal length ƒ of the combination is given by the relationship  $\frac{\mathbf{1}}{\mathbf{f}}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{f}}_{\mathbf{1}}}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{f}}_{\mathbf{2}}}$ (b) A converging meniscus lens (see Fig. 34.32a) has an index of refraction of 1.55 and radii of curvature for its surfaces of magnitudes 4.50 cm and 9.00 cm. The concave surface is placed upward and filled with carbon tetrachloride $\left(CC{I}_4\right)$ , which has n = 1.46. What is the focal length of the $\mathit{C}\mathit{C}{\mathit{I}}_{\mathbf{4}}$ -glass combination?

Step-by-Step Solution

Answer

The focal length of the combination is 8.93cm

Step 1:Relationship between object and image distance.

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

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

Step 2: Lensmaker’s formula.

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

Step 3:Proof.

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,

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

 

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.

Step 4: The focal length of the combination.

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 ${\mathbf{CCI}}_{\mathbf{4}}$ .

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

Therefore the combined focal length will be,

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