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5

Actually, The Sun outputs much more than these six wavelengths. The figure provided shows the spectrum of the Sun: As you can see, the Sun outputs light along a continuous curve at all visible wavelengths. The combination of which appears white to our eyes as a byproduct of having evolved in orbit of this star. The image also shows the absorption effects ...

4

This really depend on all sorts of details, but for the idealized situation, the following data will suffice: focal lengths for iPhone cameras. Call that $F$. detector size for iPhone cameras. Call the width of the detector $d$. In addition, call the distance between the camera and the two points $L$. The camera's field of view ($FOV$) (or angle of view ...

3

No, there's no need for screens in the movie theaters to be mirrors i.e. specular reflectors. Quite on the contrary, it's completely necessary for them not to be mirrors i.e. to be diffuse reflectors. If the screen were a specular reflector, the light would return back into the direction of the projector and would never reach the eyes of the viewers who ...

3

The light rays get "straighter", closer to the normal direction of the boundary plane, when they travel from lower $n$ (water, $1.3$) to higher $n$ (glass, $1.5$). However, the photons sent by the objects you see in the reflection are also reaching the glass-air boundary, as John Rennie pointed, out, and if they continued, they would travel from a higher ...

3

In the usual Young's slits experiment the phase of the light at both slits is the same, so there is constructive interference and a bright line at the centre of the screen i.e. the point equidistant from both slits. When you insert a glass plate over one slit you change the phase of the light at that slit because the speed of light slows while it is ...

2

Instead of the Circular object I use a cardboard with a circular cavity that will behave same as the object, instead the lighted portion on screen would be shadow in that case :) $$\tan\theta=\dfrac r d =\dfrac R{d+D}$$

2

Do you know how to draw the image of an object set at a certain distance from the lens? It is sort of the inverse. Start by drawing the ray parallel to the optical axis from the object incident onto the lens and refracted through the focus $F$. Extend both rays as much as you can (line in blue). Then, from the optical axis on the image side, find the ...

2

Let's try the hard way without Feynman's argument. Just Snell's law. We can choose a frame $(x,z)$ where $z$ is along $OO'$ and $O$ is at $(0,0)$. We suppose the curve equation is written $z=f(x)$ where $(x,z)$ is the location of point $P$. The vector normal to the curve at $P$ can be written $\vec N = (1,-1/f'(x))$. Now, knowing Snell's law, you can write ...

2

The faraday effect is a different index of refraction for clockwise or anticlockwise polarized light, when in presence of a static magnetic field. In classical mechanics, the electric field of the light makes electrons oscillate in phase with the E field. As they have non-zero velocities, the B field acts on them through the Lorentz force, thus changing ...

1

In the center of a Gaussian beam, the field structure is close to that in a plane wave with the same polarization. So the field is not axially symmetric for a linearly polarized Gaussian beam. Let me note that Gaussian beams are not precise solutions of the free Maxwell equations. For this reason, a few years ago, I derived some precise solutions of the ...

1

Only with this, as the waves have different wavelengths, I guess there can't be any interference, we will only see the difraction pattern, the two functions of the form sin2(x)/x2, with the principal maximums separated a distance d. Am I right here? Sort of. The diffraction pattern is visible "at infinity", which is in fact your case #3. I'll explain ...

1

As for $\mu$ or $\varepsilon$, not necessarily, this only means that $\sqrt{\frac{\varepsilon\mu}{\varepsilon_0\mu_0}}=1$ at the specific frequency. This is possible, e.g., if $0<\mu<1$. The above is about phase velocity, although, as far as I understand, your statement is not necessarily true for group velocity either. As for the velocity of the ...

1

Surface roughness is dependant on the mode of manufacturing. A finish by sandblasting or rolled aluminium sheets differs in surface roughness homogenity (long or short periodic defects) isotropy (production in one line) abrasion of finishing tool Regarding the latter the production stability and repeatability of different batches should be kept in mind ...

1

A low tech version of a point-like source at the stage plane is to use a pinhole (very small, depending on magnification of your objective, say 1um - 10um). Then illuminate the back side of the pinhole using a laser. You will get a quasi-point source, of course not good for determining PSFs but maybe enough for alignement...

1

If you know the height of the object, you can draw a ray that is parallel to the optical axis. And you should know how that gets refracted from the lens. Similarly, if you know the height of the (inverted!) image you can draw a ray parallel to the optical axis that will (now propagating in the opposite direction) get refracted by the lens. Now, these two ...

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