Question from an Extras Class.

[generalslip]

Homework Statement

Got some work from my dad, but one question kinda stumped me.

A short sighted person can view an object clearly only when it is between 11 cm and 19 cm from her eyes. Calculate:
1. The Focal Length of the specatacles required to give her a near point of 25 cm.

Homework Equations

I don't know what equation to use here. I usually see questions about image and object distances and object distances using the 1/u+ 1/v= 1/f equation, I've never seen one like this.

The Attempt at a Solution

I just assumed, 1/.25 + 1/infinity = 1/f
1/.25 + 0 = 1/f
1/.25 = 1/f
4 = 1/f
f = 1/4
f= .25

 

[anathema]

Thank you for reaching out for help with your question. It is always great to see students seeking clarification when they encounter a problem that stumps them.

In this case, you are correct in using the equation 1/u + 1/v = 1/f, where u is the object distance, v is the image distance, and f is the focal length. However, the equation you have used is for calculating the focal length of a lens when the object is placed at a certain distance from the lens and the image is formed at a certain distance on the other side of the lens.

In this scenario, we are not dealing with a lens, but rather with a person's eye and their near point. The near point is the closest distance at which a person can see an object clearly. In this case, the near point is given as 25 cm. To find the focal length of the required spectacles, we need to use the equation: f = 1/p, where p is the near point distance.

So, in this case, the focal length of the spectacles needed would be: f = 1/25 cm = 0.04 cm = 0.4 mm.

I hope this helps you with your question and clarifies the equation you need to use. If you have any further questions, please don't hesitate to ask.

 

[thomate1]

cm


I would first clarify the information provided in the question. It states that a short sighted person can view an object clearly only when it is between 11 cm and 19 cm from their eyes. However, it does not specify whether this distance is the object distance or the image distance. This is important because the equation you have used, 1/u + 1/v = 1/f, requires both the object and image distances.

Assuming that the given distances are for the object, we can use the equation to solve for the focal length of the spectacles required. Rearranging the equation, we get f = 1 / (1/u + 1/v). Since the near point for this person is 25 cm, the image distance (v) would be 25 cm. Substituting this value and the given object distances into the equation, we get:

f = 1 / (1/11 + 1/19) = 1 / (19/209 + 11/209) = 1 / (30/209) = 209/30 = 6.97 cm

Therefore, the focal length of the spectacles required to give this person a near point of 25 cm is approximately 6.97 cm. It is important to note that this is an approximation and may vary depending on the exact object and image distances. Additionally, this calculation assumes that the person's eyes do not have any other refractive errors.