# Light and Waves Homework

1- Give a careful description comparing and contrasting the interference pattern of light passing through a double slit a arrangement for blue light and for a green light , be sure to justify the compartment carefully ??

In all cases when coherent monochromatic light is subjected to a double slit diffraction experiment, an interference pattern is formed from bright and dark fringes form on a screen. Always on the center of the screen there is the brightest fringe, followed by a dark fringe, a second order bright fringe, and so on. The bright fringes represent the geometric place on the screen where the difference in path between the two diffracted light rays undergo constructive interference whereas the dark fringes represent the place where destructive interference takes place.

For different wavelengths of light there are also differences. The main difference is that for for longer wavelengths the angular (and linear) separation on the screen is higher between fringes. The equation that describes this is

d*sin(theta) = k*lambda

where theta is the angular separation, k an integer and lambda is the wavelength.

(Greater lambda means greater theta)

For linear separation sin(theta) =tan(theta) = D/L

(D is separation and L the distance to screen)

D will also b higher for higher lambda.

2 - The opening to a cave is a tall, 30-cm-wide crack. A bat that is preparing to leave the cave emits a 30 kHz ultrasonic chirp. Do you expect much diffraction of the second wave as it exits the cave ? Explain

Diffraction of waves takes place when a wave in its straight trajectory encounters an obstacle and “goes around this obstacle”. This phenomena is only characteristic to waves and for it to happen it needs that the wavelength is comparable in length with the obstacle dimension. For longer wavelengths than the dimension of the obstacle the diffraction does not take place, whereas for shorter wavelengths the diffraction is pronounced.

For the bat ultrasonic chirp the wavelength is lambda = V*T = V/F =331/30000 = 0.011 m = 11cm. Since wavelength of bat chirp is comparable with the crack size, (tens of centimeters) there all be some diffraction in of the sound. Pronounced diffraction would take place if the bat sound wavelength would have been of about 1 cm.

3- Explain how an optical fiber is constructed and how it functions ?

An optical fiber is constructed out of a core material and a coating material. The core material has a higher index of refraction than the coating material. Because of the fact that the core material has a greater index of refraction than the coating, for small angles with the wall of the incident light the light is totally reflected back into the fiber. Therefore only a small fraction of the light inside is lost .

The total internal reflection takes place for angles of incidence (measured with respect to the wall normal) higher than sin(i)> 1/n where n is the relative index of refraction of core with respect to coating (n = n1/n2)

4 - An inexpensive children telescope has an objective lens with focal length of 30 cm and diameter of 2 cm. can the telescope theoretical resolve the two headlights on a car that 6,000 m away ? use 1 m for the separation of the headlights and 500 nm for the wavelength light.

Two rays are said that are resolved by a lens if the angle theta between them is bigger than

$sin(theta) = 1.22*lambda/D$

where lambda is the light wavelength and D is the lens diameter.

In the case of the telescope in the problem $sin(theta) = 1.22*500*10^{-9}/0.02 =3.05*10^{-5}$

$theta = 1.75*10^{-3} degree =3.05*10^{-5} rad$

The angle between incoming rays of light from car headlights is

$tan(alpha) = d/L = 1/6000 =1.67*10^{-4}$

(d is the separation of headlights and L the distance to the car)

$alpha = 1.67*10^{-4} rad$

since the telescope angular resolution is smaller than the angle between headlights than YES the telescope can resolve the two points.

5- Explain how a microscope works. include a clear labeled diagram ?

A simple microscope is mainly formed by two lenses. An eyepiece and and objective. The objective is a lens with a high diameter (to allow as much as possible light to enter it) and with magnifications between 5x to 100x. This lens forms a real and inverted image Y2 of the object Y1. The eyepiece is a second lens mounted in such a way that its focal length (Foc) nearly coincides with the position of the image formed by the ocular (in reality its focal length is slightly further than the position of the image formed by the ocular) . This way the ocular is forming a virtual image Y3 almost at infinity, and because of this in tum the image formed Y3 will be very big as compared to y2.

The magnification of eyepiece is between 2x and 50x and because the distance (e) between $F_{oc}$ and $F_{ob}$ is very small, the total magnification of the microscope is about

$M = F_{oc}*F_{ob}$