Final answer:
The height of the image of the print on the retina when the book is held 30.0 cm from the eye is approximately 1.05 mm. Therefore, the correct option is a.
Step-by-step explanation:
To determine the height of the image on the retina, we can use the magnification formula:
![\[ \text{Magnification} = \frac{\text{Image height}}{\text{Object height}} \]](https://img.qammunity.org/2024/formulas/physics/high-school/woczngllotn03dutyu7celcbngtlpws0rw.png)
Given that the book is held 30.0 cm from the eye and the print height is 3.50 mm, we can use the magnification formula to find the image height on the retina. The magnification is also related to the object distance (distance between the print and the eye) and the image distance (distance from the lens to the retina) by the lens formula:
![\[ \text{Magnification} = \frac{\text{Image distance}}{\text{Object distance}} \]](https://img.qammunity.org/2024/formulas/physics/high-school/gtnf44q8u10olx5iap63vutc7zu3gh55ew.png)
By combining these two formulas, we get:
![\[ \frac{\text{Image height}}{\text{Object height}} = \frac{\text{Image distance}}{\text{Object distance}} \]](https://img.qammunity.org/2024/formulas/physics/high-school/ly0flkgjd31xwchs0lbh4iw95ylnae09mr.png)
Rearranging for the image height, we have:
![\[ \text{Image height} = \frac{\text{Object height} * \text{Image distance}}{\text{Object distance}} \]](https://img.qammunity.org/2024/formulas/physics/high-school/ulf5exifcalktg83q2h4o1raiq8nn45auq.png)
Substituting the given values, we get:
![\[ \text{Image height} = \frac{3.50 \, \text{mm} * 30.0 \, \text{cm}}{30.0 \, \text{cm}} = 1.05 \, \text{mm} \]](https://img.qammunity.org/2024/formulas/physics/high-school/ifk1lqi19cxytxgazo5rwv9abqm5tkbw5j.png)
Therefore, the image of the print on the retina is approximately 1.05 mm, corresponding to option (a).