Recent content by Petra de Ruyter

Greetings Many thanks willem2. In my book it is (1.19). I appreciate all your help on this. I won't be doing this unit yet, but I wanted to get a head start on it. cheers

Greetings the source is in the book "The Light Fantastic: A modern introduction to Classical and Quantum Optics. The answer to the question is listed as 1.06 kVm^-1 but does not state how this is calculated, I need some guidance as to how this figure was reached?

Is there a formula? A few websites state that it is 1.06 kVm^-1, however they do not say how it is calculated.

Hi there Only known variables include the solar flux on the Earths atmosphere at the equator is 1.5 kWm^ -2. I've calculated the total energy from the Sun as 4 x 10^26 W, but can't seem to get to a calc of the mag of electric field at equator (measured in kVm^-2).

Hi there This is the question "What is the object distance when a myopic person with a far point of 150 cm, can see near images clearly at 20 cm from the eye." I keep thinking that I need more information.

Homework Statement A myopic person has a far point of 150cm and can see objects clearly at 20cm near point. What is the object distance? Homework Equations Thin lens equation 1/f =1/d1-do. The Attempt at a Solution di = -150cm because it's on the left of the lens. do = should =...

Many thanks to all that have posted.

This syringe is filled with air only and connected to another syringe that is empty. I'm looking for a text reading that will help me understand what is happening, I don't necessary need anything solved just need a good resource that I can refer to.

Hi all I'm looking for general information on "back pressure" specially in a syringe. So when you put your finger over the top of a syringe and depress the plunger then release the pressure on the plunger and the plunger moves back slightly. I'm struggling to find some specific information...

Hi there I have a general question. I understand the process of light through a optically denser material, slower speed etc... but why would a laser beam in oil be easier to "see" than a laser beam through water. Cheers

Hi there. I could not measure it because I did not have the refractive index, now I do and all is well. many thanks for all your assistance.

53.00 and 51.98 The laser that was provided was pretty weak and the angles are a bit sketchy. Cheers

Air and water :-)

Hi all Tearing my hair out trying to calculate the Maximum acceptance angle for a problem. Experiment involved shining a laser into a tube of water. I have calculated the angle of refraction based on the average of two critical angles (one known and one measured manually) as 52.49 degrees...

Just so that I understand. You should attempt to have the dep variable on the left and the indep variable on the right before classification? Cheers Petra