Get Help with Light and Optics Physics and Ace Your Tests

Does sound travel faster than light?

Suppose the distance from the lens system of the eye (cornea + lens) to the retina is 16 mm.
(a) What must the power of the lens be when looking at distant objects?
(b) What must the power of the lens be when looking at an object 18.2 cm from the eye?
(c) Suppose that the eye is farsighted; the person cannot see clearly objects that are closer than 1.0 m. Find the power of the contact lens you would prescribe so that objects as close as 18.2 cm can be seen clearly.

Find the distributions of the random variables that have each of the following moment generating functions:
$m(t)=\frac{e^t}{2-<!--\; -\; -->e^t}$

What is the refractive index evaluated in?

How can you identify whether a picture is upright or upside down?

A measure of an angle 240∘ is __?

Photons of minimum energy 496 kJ/mol are needed to ionize sodium atoms. Calculate the lowest frequecy of light that will ionize a sodium atom. A1.24×1014 s−1 B1.24×1015 s−1 C2.48×1015 s−1 D2.48×1014 s−1

State the factor(s) that needed to be considered while planning a single slit diffraction experiment, in order to make a diffraction pattern easily visible.

Is Heisenberg's Uncertainty Principle applicable to light?

What is the experimental evidence that light is an electromagnetic wave?

I'm having trouble obtaining this formula.
Considering the curvature of the Earth (R is the Earth radius) and a non-vertical direction (zenith angle $\theta$), the relation between h and path length L in the atmosphere is:
$h=L\cos <!--\; \; -->\theta +\frac{1}{2}\frac{L^2}{R}{\sin }^2<!--\; \; -->\theta$
h is the atmosphere's height.
I understand the first term (which due to the inclination) but I can't find a way to get the second term (which is introduced by considering the "roundness" of the Earth)
Any help would be greatly appreciated

Suppose you are measuring double-slit interference patterns using an optics kit that contains the following options that you can mix and match: a red laser or a green laser, a slit width of 0.04 or 0.08 mm; a slit separation of 0.25 or 0.50 mm. Estimate the minimum distance L you can place a screen from the double slit that will give you an interference pattern on the screen that you can accurately measure using an ordinary 30 cm (12 in) ruler.

Why do electrons on the surfaces of a lens focus photons instead of scattering them in all directions?

When parallel rays are refracted/reflected to a point in ray diagrams, we say an image is formed. However practically the real image of a star/celestial body will not be an infinitesimally small point. Is this a breakdown of the model we use as the rays won't be perfectly parallel?

Can a broken light filter shift the wavelength of light exiting it?

In all consulted literature, the transfer function of the free space is given as follows:
$\exp <!--\; \; -->(-<!--\; -\; -->i{k}_{z}d)=\exp <!--\; \; -->(-<!--\; -\; -->i2\mathrm{\pi <!--\; \pi \; -->}d\sqrt{1/{\mathrm{\lambda <!--\; \lambda \; -->}}^2-<!--\; -\; -->{\mathrm{\nu <!--\; \nu \; -->}}_x^2-<!--\; -\; -->{\mathrm{\nu <!--\; \nu \; -->}}_y^2})$
When referring to Fourier Optics, they derive the transfer function from the following equation:
$H({\mathrm{\nu <!--\; \nu \; -->}}_x,{\mathrm{\nu <!--\; \nu \; -->}}_y)=\frac{f_{in}(x,y)}{f_{out}(x,y)}$
I'm wondering why they do it this way. I thought the transfer function is defined in the frequency domain rather than in the time domain (frequency and space for the fourier optics respectively).

How is the Rayleigh criterion connected to the Abbe limit?

Why the incessant confusion between diffraction and interference?

Light incident normally on a grating of total ruled width of $5.1\times <!--\; \times \; -->{10}^{-<!--\; -\; -->3}\text{ m}$ with 2520 lines in all. Find out if two sodium lines with wavelengths $5890{\text{ A}}^{\circ <!--\; \circ \; -->}$ and $5896{\text{ A}}^{\circ <!--\; \circ \; -->}$ can be seen distinctly by the grating.

Recently I've heard that you can't heat up stuff beyond sun's surface temperature using sun+lens combination. It is because you can't transport heat from colder object to a hotter object.
Of course, the bigger lens you use, the closer the heated object gets to the sun's surface temperature. So there's a certain lens size, beyond which you can't get heated object any hotter.
But when I exceed that lens' size... where does the extra heat goes? The lens collects solar power from its entire surface area, and focuses it in one spot. So if you reach certain size you collect lots of solar power, but the heated object doesn't increase the temperature. So where all the extra energy goes?