Chapter 18, Problem 67A

(a)

To determine

The position and height of the game piece.

Answer to Problem 67A

  $d_o=7.5\times {10}^{-2}\text{ m}$

  $h_o=3.00\times {\text{10}}^{-2}\text{ m}$

Explanation of Solution

Given:

Height of an image is $h_i=2.0\text{ cm = 2}\times {\text{10}}^{-2}\text{ m}$

Distance between the image and the diverging lens is $d_i=-5.0\text{ cm = - 0}\text{.05 m}$

Focal length of a diverging lens is $f=-15.0\text{ cm = - 0}\text{.15 m}$

Formula used:

The relationship between the object distance $\left(d_o\right)$ , image distance $\left(d_i\right)$ and the focal length $\left(f\right)$ of a lens is given by,

  $\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}$

  $\Rightarrow \frac{1}{d_o}=\frac{1}{f}-\frac{1}{d_i}$

  $\Rightarrow \frac{1}{d_o}=\frac{d_i-f}{f{d}_i}$

  $\Rightarrow d_o=\frac{f{d}_i}{d_i-f}$ $\left(1\right)$

The magnification $\left(M\right)$ equation gives the relationship between the image height $\left(h_i\right)$ , object height $\left(h_o\right)$ , image distance $\left(d_i\right)$ , and object distance $\left(d_o\right)$ as,

  $M=\frac{h_i}{h_o}=-\frac{d_i}{d_o}$

  $\Rightarrow \frac{h_i}{h_o}=-\frac{d_i}{d_o}$

  $\Rightarrow h_o=-\frac{d_o}{d_i}{h}_i$ $\left(2\right)$

Calculation:

Position of the game piece can be calculated by substituting the numerical values in equation $\left(1\right)$ ,

  $d_o=\frac{\left(\text{- 0}\text{.15 m}\right)\left(\text{- 0}\text{.05 m}\right)}{\left(\text{- 0}\text{.05 m}\right)-\left(\text{- 0}\text{.15 m}\right)}$

  $=\frac{7.5\times {10}^{-3}{\text{ m}}^2}{0.1\text{ m}}=7.5\times {10}^{-2}\text{ m = 7}\text{.5 cm}$

Height of the game piece can be calculated by substituting the numerical values in the equation $\left(2\right)$ ,

  $h_o=-\frac{\left(7.5\times {10}^{-2}\text{ m}\right)}{\left(\text{- 0}\text{.05 m}\right)}\left(\text{2}\times {\text{10}}^{-2}\text{ m}\right)$

  $=\frac{15\times {10}^{-4}{\text{ m}}^2}{\text{0}\text{.05 m}}=3.00\times {\text{10}}^{-2}\text{ m}=3.00\text{ cm}$

Conclusion:

Position of the game piece is at $7.5\times {10}^{-2}\text{ m}$ from the lens and the height of the game piece is $3.00\times {\text{10}}^{-2}\text{ m}$ .

(b)

To determine

The position, height, and orientation of the image.

To identify: Whether the image formed is real or virtual.

Answer to Problem 67A

  $d_{i,new}=-15\times {10}^{-2}\text{ m}$

  $h_{i,new}=6.0\times {10}^{-2}\text{ m}$

The image is upright and virtual.

Explanation of Solution

Given:

Since diverging lens is replaced by a converging lens,

Focal length of the lens, $f_{new}=-f=-\left(-15.0\right)\text{ cm = 15 cm = 0}\text{.15 m}$

Height of the game piece, $h_o=3.00\text{ cm}=3.00\times {\text{10}}^{-2}\text{ m}$

The distance between the game piece and the lens, $d_o=\text{7}\text{.5 cm}=7.5\times {10}^{-2}\text{ m}$

Formula used:

The relationship between the object distance $\left(d_o\right)$ , image distance $\left(d_i{{}_{,}}_{new}\right)$ and the focal length $\left(f_{new}\right)$ of a lens is given by,

  $\frac{1}{f_{new}}=\frac{1}{d_o}+\frac{1}{d_i{}_{,new}}$

  $\Rightarrow \frac{1}{d_{i,new}}=\frac{1}{f_{new}}-\frac{1}{d_o}$

  $\Rightarrow \frac{1}{d_{i,new}}=\frac{d_o-f_{new}}{f_{new}{d}_o}$

  $\Rightarrow d_{i,new}=\frac{f_{new}{d}_o}{d_o-f_{new}}$ $\left(3\right)$

The relationship between the image height $\left(h_{i,new}\right)$ , object height $\left(h_o\right)$ , image distance $\left(d_{i,new}\right)$ , and object distance $\left(d_o\right)$ as,

  $\frac{h_{i,new}}{h_o}=-\frac{d_{i,new}}{d_o}$

  $\Rightarrow h_{i,new}=-\frac{d_{i,new}}{d_o}{h}_o$ $\left(4\right)$

Calculation:

The distance between the lens and the image can be calculated by substituting the numerical values in equation $\left(3\right)$ ,

  $d_{i,new}=\frac{\left(\text{0}\text{.15 m}\right)\left(7.5\times {10}^{-2}\text{ m}\right)}{\left(7.5\times {10}^{-2}\text{ m}\right)-\left(\text{0}\text{.15 m}\right)}$

  $=\frac{1.125\times {10}^{-2}{\text{ m}}^2}{-0.075\text{ m}}=-15\times {10}^{-2}\text{ m}=-15\text{ cm}$

The height of the image can be calculated by substituting the numerical values in equation $\left(4\right)$ ,

  $h_{i,new}=-\frac{\left(-15\times {10}^{-2}\text{ m}\right)}{\left(7.5\times {10}^{-2}\text{ m}\right)}\left(3.00\times {\text{10}}^{-2}\text{ m}\right)$

  $=6.0\times {10}^{-2}\text{ m}=6.0\text{ cm}$

The values of $d_{i,new}$ and $h_{i,new}$ indicates that the image formed is virtual and upright.

Conclusion:

The distance between the lens and the image is $-15\times {10}^{-2}\text{ m}$ .

The height of the image is $6.0\times {10}^{-2}\text{ m}$ .

The image formed is virtual and upright.