1. a Ray of Blue Light Inside a Rectangular Block of Transparent Material Is Totally Internally

PHYS 262 Spring 2007 Exam 2 Name ______April 3, 2007 Please Print

Problem 1 ______Problem 2 ______Problem 3 ______

TOTAL: ______

1/ f  1/ d o 1/ di f  r / 2 m  h'/ h  di / d o n  c / v sin c  n2 / n1

n1 sin1  n2 sin 2 m   '/ m  25/ f m  25ℓ/ f e f o m   f o / f e m  d sin (m  0.5)  d sin P  1/ f 2   2r /  1/ f  (n 1)(1/ R1 1/ R2 ) I  I 0 cos  / 2 2 n   / n I  I 0 cos  sin c  n2 / n1 1. A ray of blue light inside a rectangular block of transparent material is totally internally reflected at point A but not at B where it emerges into the air – see figure.

(a) Find the minimum value of n for the block from the information provided. A B 600 sin c  n2 / n1  1/ n  sin 60  0.866 n  1/ 0.866  1.15

(b) Find the maximum value of n for the block from the information provided

sin c  n2 / n1  1/ n  sin 30  0.5 n > 1 / 0.5 = 2.0

(c) Now suppose the ray were red light instead of blue. Would it necessarily be totally internally reflected at point A? Explain your answer.

No. n for red light is less tha n for blue, implying that the critical angle for red must be larger than blue, therefore although the blue ray exceeds the critical angle, the red ray might not.

(d) In one clear English sentence explain under what conditions the ray approximation applies to light. Also explain why we rarely hear about “sound rays.”

The approximation holds as long as the wavelength is small compared to the size of any opening or obstacles the wave encounters. When it is untrue as in the case of sound (which has a fairly large wavelength compared to many objects the sound encounters), the ray approximation doesn’t hold, and diffraction becomes very important. 2. (a) Construct a neat ray diagram that shows how when someone views a distant object with a telescope angular magnification is achieved. (Hint: Be sure your diagram has fo>> fe)

(b) When an object is placed 10 cm in front of a certain lens an inverted image is produced that is twice as large as the object. Find the focal length of this lens. do =10 cm, m = -2 = -di/do = -di/10, so di = 20 cm

1/f = 1/di + 1/do, solving for f yields f = + 6.67 cm

(c) I am so nearsighted that I cannot see things clearly that are further than 4 inches from my face. What power eyeglasses should be prescribed for me?

Use: di = -4 in = -0.102 m and do = infinity, solving the lens eqn yields P = 1/f = -9.8 Diopters

(d) An object that is 10 cm in front of a lens produces an image on a screen that is 30 cm behind the lens. How far must the lens be moved toward the screen to again yield an image falling on the screen? Assume the object and screen locations are fixed.

Moving screen by 20 cm toward screen basically interchanges the image & object distances & therefore still satisfies the lens eqn. 3 (a) Microwaves having a 1.3 cm wavelength are incident on a pair of slits separated by 7 cm. How many maxima will be produced in the interference pattern? (Count maxima at positive, negative and zero angles.) Under what conditions will the maxima all be about the same height (or intensity)?

m  d sin m(1.3)  7.0sin 90 m = 5.4 Max integer is 5, which gives 11 maxima: m = -5, -4, -3, -2, -1 , 0, 1, 2, 3, 4, 5 (this part was 6 pts)

The maxima are all about the same height only when the slits are all very narrow, so that the diffraction pattern spreads out to all angles equally (this part was 4 pts)

(b) It is found that when microwaves of another wavelength are passed through these same slits their third minima falls at the same angle that the first wavelength had its first maxima (i.e., first away from zero). What is the new wavelength?

(m'0.5)' d sin  m where m’ = 2 & m = 1. Solving for '  / 2.5  0.52cm

(c) A and B represent two point sources of microwaves that are produced in phase with each other. It is found that when the waves from each source reach point C they destructively interfere. What are all possible values of the wavelength based on this information? C r  10  6  (m 1/ 2)

so   4 /(m 1/ 2) cm where m = 0,1,2,3,4,… 6 cm m  d sin not used here Since r  d sin A B 8 cm (d) An air wedge is formed when a piece of paper is inserted to separate two plates of glass at one edge. Light of 500 nm wavelength is incident normally from above and interference fringes are created. If the plates are 5 cm long, and the paper is 0.1 mm thick how many bright fringes are seen on the plate?

2t = m  gives m = 400 m  d sin not used here