A2 Health Physics Eye question

[ejonesdj]

Hello everyone,
Really struggling to answer a practice exam question on the eye, and was wandering if anyone could help, here's the question:

"The unaided eye of a student has a power of 59.0D when viewing an object at infinity. When using her spectacle lens, the near point is 25cm from her eye.

Calculate the distance of the near point for her unaided eye.
The power of the student's glasses lens is 1.5D"

I do have the markscheme however I'm very confused as to how they have got their answer.

Thanks in advance,
Ed

 

[ejonesdj]

I understand the first part of the markscheme - where the 1/f=1/u +1/v formula (where f= focal length, u= object-lens distance and v= image-lens distance) which gives the object-lens distance by substituting 59D as the power (1/f).
So:
59D = 1/infinity +1/u, thus giving u= 1.69cm

The next part of the markscheme confuses me:
It says that you need to work out the power of the lens and eye together by doing the following:
Pe + 1.5 = 1/0.25 + 1/0.017
Pe = 61.5D

Im struggling to see how that works...
The markscheme continues to give the final answer stating:
61.5 = 1/u + 59
giving u = 40cm
Im also not sure what's going on here..
If someone could explain what there doing that would be great,
thanks in advance
Ed

 

[Moetasim]

Hello Ed,

I can understand your confusion with this question. Let me try to explain the solution to you.

Firstly, we need to understand what the terms "power" and "near point" mean in this context. The power of a lens is a measure of its ability to bend light. It is measured in diopters (D). The near point is the closest distance at which the eye can focus on an object.

Now, let's look at the given information. The unaided eye has a power of 59.0D when viewing an object at infinity. This means that the unaided eye has a natural ability to focus on objects at a distance of infinity. However, when the student uses her glasses, the near point becomes 25cm from her eye. This means that with the glasses, the student's eye can only focus on objects at a distance of 25cm.

To calculate the distance of the near point for the unaided eye, we can use the formula:

Power = 1/Distance

Substituting the values given, we get:

59.0D = 1/Distance

Solving for Distance, we get:

Distance = 1/59.0D = 0.0169m = 16.9cm

So, the distance of the near point for the unaided eye is 16.9cm.

Now, to calculate the power of the glasses lens, we can use the formula:

Power = 1/Distance

Substituting the values given, we get:

1.5D = 1/0.25m

Solving for Distance, we get:

Distance = 1/1.5D = 0.6667m = 66.67cm

So, the power of the glasses lens is 66.67D.

I hope this explanation helps you understand the solution better. If you have any further questions, please feel free to ask. Best of luck with your exam!