Chapter 24, Problem 88GP

To determine

The locations of the red and yellow images formed by the lens.

Answer to Problem 88GP

  $\text{79}\text{.4}\text{cm}\text{\&}\text{75}\text{.5}\text{cm}$

Explanation of Solution

Given info:

Refraction index of red light, ${\text{n}}_{\text{red}}=1.5106$

Refraction index of yellow light, ${\text{n}}_{\text{yellow}}=1.5226$

Frequency, $f=6.0\times {10}^{6}Hz$

Formula used:

  $\frac{1}{\text{f}}=\left(\text{n}-1\right)\left(\frac{1}{{\text{R}}_1}+\frac{1}{{\text{R}}_2}\right)$ and $\frac{1}{{\text{d}}_0}+\frac{1}{{\text{d}}_{\text{i}}}=\frac{1}{\text{f}}$

Where,

  $f$ is focal length.

  ${\text{d}}_0$ is object distance.

  ${\text{d}}_{\text{i}}$ is image distance.

  $n$ is refractive index.

  $\text{R}$ is radius of curvature,

Calculation:

Using the lens maker’s equation to find the two focal lengths, $\frac{1}{\text{f}}=\left(\text{n}-1\right)\left(\frac{1}{{\text{R}}_1}+\frac{1}{{\text{R}}_2}\right)$

  $\because {\text{R}}_{\text{1}}=\infty \&{\text{R}}_2=18.4\text{cm}$

  $\therefore \text{f}=\frac{18.4\text{cm}}{\text{n}-1}$

  $\therefore {\text{f}}_{\text{red}}=\frac{18.4\text{cm}}{1.5106-1}=36.04\text{cm}$ and ${\text{f}}_{\text{yellow}}=\frac{18.4\text{cm}}{1.5226-1}=35.21\text{cm}$

Now, find the two-image distance, $\frac{1}{{\text{d}}_0}+\frac{1}{{\text{d}}_{\text{i}}}=\frac{1}{\text{f}}$ ,

  $\therefore \frac{1}{66.0\text{cm}}+\frac{1}{{\text{d}}_{\text{i-red}}}=\frac{1}{36.04\text{cm}}$

  $\therefore {\text{d}}_{\text{i-red}}=79.4\text{cm}$

Similarly for yellow light,

  $\therefore \frac{1}{66.0\text{cm}}+\frac{1}{{\text{d}}_{\text{i-yellow}}}=\frac{1}{35.21\text{cm}}$

  $\therefore {\text{d}}_{\text{i-yellow}}=75.5\text{cm}$

Conclusion: The locations of the red and yellow images formed by the lens will be at distance of $\text{79}\text{.4}\text{cm}\text{\&}\text{75}\text{.5}\text{cm}$ respectively.