Chapter 25, Problem 63GP

To determine

Maximum magnification that the child can obtain.

Also, the magnification that a normal eye can obtain.

Answer to Problem 63GP

Maximum magnification value is $\text{2}\text{.9}\text{x}$ and normal magnification value is $\text{4}\text{.1}\text{x}$ .

Explanation of Solution

Given info:

First near point, ${\text{N}}_1\text{=15}\text{cm}$

Second near point, ${\text{N}}_2\text{=25}\text{cm}$

Focal length, $\text{f}\text{=}\text{8}\text{.0}\text{cm}$ .

Formula used:

It is known that $\text{M}\text{=}\frac{\text{N}}{\text{f}}$ . Where eye focused at $\infty$ and $\text{N = 25}\text{cm}$ for a normal eye. Here, $\text{M}$ is the magnifying power and $\text{N}$ is the object at the near point, of eye, $\text{f}$ is the focal length.

The magnification when the eye is focused at its near point $\left(\text{N = 25}\text{cm}\right)$ , and also, the lens is near the eye:

  $\text{M}\text{=1}\text{+}\frac{\text{N}}{\text{f}}$

Calculation:

The maximum magnification is achieved with the image at the near point:

  $\text{M}\text{=1}\text{+}\frac{{\text{N}}_1}{\text{f}}$

  $\text{M}\text{=1}\text{+}\frac{\left(\text{15}\text{.0}\text{cm}\right)}{\text{8}\text{.0}\text{cm}}\text{=}\text{2}\text{.9}\text{x}$

For an adult we have

  $\text{M}\text{=1}\text{+}\frac{{\text{N}}_2}{\text{f}}$

  $\text{M}\text{=1}\text{+}\frac{\left(\text{25}\text{.0}\text{cm}\right)}{\text{8}\text{.0}\text{cm}}\text{=}\text{4}\text{.1}\text{x}$

Conclusion:

The required values are $\text{2}\text{.9}\text{x}$ and $\text{4}\text{.1}\text{x}$ .