# Aperture Question

1. Feb 9, 2016

### toesockshoe

1. The problem statement, all variables and given/known data

A standard 14.16-inch (0.360-meter) computer monitor is 1024 pixels wide and 768 pixels tall. Each pixel is a square approximately 281micrometers on each side. Up close, you can see the individual pixels, but from a distance they appear to blend together and form the image on the screen.

If the maximum distance between the screen and your eyes at which you can just barely resolve two adjacent pixels is 1.30 meters, what is the effective diameter d of your pupil? Assume that the resolvability is diffraction-limited. Furthermore, use λ=550 nanometers as a characteristic optical wavelength.

2. Relevant equations

$$sin(\theta) = \frac{1.22* \lambda}{D}$$
3. The attempt at a solution

using small angle formula, you can assume $$sin(\theta) = \theta$$ So, $$D = \frac {1.22 \lambda}{\theta}$$. $$\theta = \frac{281}{10^6}$$ .. AND $$\lambda = \frac{550}{10^9}$$. Plugging all these values in, I get D=2.387 mm which is incorrect according to the homework checker. Where am I going wrong?

2. Feb 9, 2016

### TSny

Hello. Can you explain why the above expression represents θ? If you were to include units with your numbers, would you get the correct units for θ?

3. Feb 9, 2016

### toesockshoe

It is 281micrometers/10^6 micrometers. This would be dimensionless which is a valid argument for theta.

4. Feb 9, 2016

### toesockshoe

It is equivalently 281 times 10^-6 m/1m.

5. Feb 9, 2016

### TSny

Why the 1 m in the denominator?

6. Feb 9, 2016

### toesockshoe

should it be 1.3m in the denominator instead? because tantheta is approx sin theta for small angles? ... i still dont get right answer.

7. Feb 9, 2016

### TSny

Forget the numbers for a moment. Let's make sure we understand the concept. Can you describe in words how to find θ?

8. Feb 9, 2016

### toesockshoe

yeah... you can use tan^-1(opposite/adjacent). opp is the distance between 2 pixels, and adj distance is the distance from you to the screen (which is 1.3). we have the opp distance and adj distance so we can find theta. units, cancel out because you are dividing meters/meters... and you have an acceptable angle measure.

9. Feb 9, 2016

### TSny

OK, that will work for small angles. For small angles you can go on and approximate d/L by θ (in radians), where d is distance between pixels and L is distance from eye to pixels. So, yes, you should be dividing by 1.3 m.

I'm not sure why you are not getting the right answer now. What are you getting for D?

Last edited: Feb 9, 2016
10. Feb 9, 2016

### toesockshoe

I got it. thank you.

11. Feb 9, 2016

Good work.