Question

A student’s eyes, while reading the blackboard, have a power of 51.0 D. How far is the board from his eyes?

a) 0.196 m
b) 0.250 m
c) 0.185 m
d) 0.222 m

Answer 1

Using the formula for lens power, P = 1/f, the distance from the student's eyes to the blackboard with a power of 51.0 D is calculated to be approximately 0.0196 meters, or 19.6 centimeters. This number seems unrealistic for actual blackboard reading distances, suggesting it is a theoretical scenario.

Step-by-step explanation:

To determine how far the blackboard is from the student's eyes given a power of 51.0 diopters (D), we need to use the formula for the power of a lens, which is P = 1/f, where P is the power in diopters and f is the focal length in meters. To solve for the focal length (distance from the eyes to the blackboard), we rearrange the formula to f = 1/P.

Consequently, substituting the value of P = 51.0 D into the formula, we get f = 1/51.0, which equals approximately 0.0196 meters, or 19.6 centimeters.

Therefore, the distance from the student's eyes to the blackboard is about 19.6 centimeters, which would be highly unusual in a real-world scenario, indicating that the student is using an accommodative reflex to read at this short distance, or it could be an error in the problem statement. Nonetheless, for this problem, we'll use these values as given.