PHYSICS III
Solution to Exam 4

Monday 5/16/16	Name
O205: Periods 1, 2
JFW	CM:
	Period:

	(1)		(25)
	(2)		(25)
	(3)		(25)
	(4)		(25)
	TOTAL		(100)

1. A transverse upright object is placed somewhere between C and F of a concave mirror resulting in a magnified image three times larger than the object. The radius of curvature of the mirror is 24.0 cm.
   
   (a) [2 pts] In the figure below, indicate, qualitatively, where points C and F should be placed relative to the vertex.
   ANS: First place C on the optical axis. Then place F halfway between C and V.
   (b) [5 pts] Calculate the focal length of the mirror.
   
   

   f=R/2=+24.0 cm/2=+12.0 cm

   
   (c) [10 pts] Calculate the position of the object.
   
   

   s¢=3s

   
   

   1/s+1/s¢=1/f

   
   

   1/s+1/3s=1/f

   
   

   s=4f/3=4(+12.0 cm)/3=+16.0 cm

   
   (d) [3 pts] Calculate the position of the image.
   
   

   s¢=3s=3(+16 cm)=+48.0 cm

   Note: the image is real, inverted, and, as stated, magnified.
   (e) [5 pts] Using two principal rays, draw a ray diagram indicating the position of the image.
2. Two thin, parallel lenses are separated by a distance d. The left lens is diverging with a focal length f_d. The other lens is converging with a focal length f_c. An upright, transverse object of height y is located a distance s₁ to the left of the diverging lens.
   

   fd=-10.0 cm d=10/3 cm s1=20.0 cm y=1.0 cm

   
   (a) [6 pts] If the diverging lens is equi-concave with an index of refraction of n=1.60, calculate the radius of curvature of each surface.
   
   

   1 fd = (n-1) æ è 1 R1 - 1 R2 ö ø

   
   

   R1=-R R2=+R

   
   

   1 fd = (n-1) æ è - 1 R - 1 R ö ø R = -2(n-1)fd = -2(1.60-1)(-10.0 cm) = 12.0 cm

   
   (b) [5 pts] Calculate the position of the image produced by the diverging lens.
   
   

   1 s1 + 1 s¢1 = 1 fd 1 +20.0 + 1 s¢1 = 1 -10.0 s¢1 = -6.67 cm

   
   (c) [6 pts] How tall is the image? Is the image real or virtual? Is the image upright or inverted?
   
   

   m=-s¢1/s1=-(-6.67 cm)/20.0 cm=+0.333

   
   

   y¢1=my1=0.333(1.00 cm)=0.333 cm

   The first image is virtual, upright and, demagnified.
   (d) [8 pts] The final image, as produced by the converging lens is observed to be at infinity. Calculate the focal length of the converging lens, f_c.
   

   I1(real)=O2(real) ® s2=d+|s¢1|

   
   

   s2 = 3.33 cm+6.67 cm = 10.0 cm

   
   

   1 s2 + 1 s¢2 = 1 fc 1 +10.0 + 1 ¥ = 1 fc fc = +10.0 cm
3. The rays of monochromatic plane wavefronts fall on a double slit system at normal incidence. The slits are identical and separated by a distance d. The slits behave as two coherent sources emitting in phase. A parallel viewing screen is a distance R away. The numerical values are
   

   l = 540 nm d=80.0 mm R=1.50 m

   
   (a) [10 pts] Calculate the distance from the center of the viewing screen to the m=3 bright fringe.
   
   

   dsinq = ml (80.0 mm)sinq3 = 3(0.540 mm) q3 = 1.16o y3 = Rtanq3 = (150 cm)tan1.16o = 3.04 cm

   
   (b) [5 pts] What is the phase difference between the two rays arriving at the m=3 bright fringe and the two rays arriving at the m=5 bright fringe?
   ANS: 4p rad
   (c) [10 pts] An observer measures an intensity I_o at the center of the image screen (y=0). The observer moves away from the center point until there is a 50% drop in the intensity. Calculate the phase difference between the two rays arriving at this point and what is the corresponding path difference between the two rays?
   The phase difference is
   

   Io 2 = Iocos2 (f/2) cos2 (f/2) = 1/2 cos(f/2) = 0.707 f = p/2 rad ® 90o

   The path difference is
   

   f = kDr Dr = fl 2p = æ è p 2 ö ø æ è 540 nm 2p ö ø = 135 nm

   This corresponds to an angle of q_1/2=0.097^o which is found from
   

   Dr=dsinq
4. The rays of monochromatic plane wavefronts fall at normal incidence on a screen containing a single slit of width a. A parallel viewing screen is a distance R away. The numerical values are
   

   a=10.0 mm l = 620. nm R=1.20 m

   
   (a) [8 pts] Determine the location of the first intensity minimum. Give the location in terms of the distance from the center point on the viewing screen.
   
   

   asinq = nl (10.0 mm)sinq1 = (1)(0.620 mm) q1 = 3.555o y1 = Rtanq1 = (120. cm)tan(3.555o) = 7.45 cm

   
   (b) [5 pts] Calculate the width of the central bright fringe.
   
   

   DD1 = 2y2=2(7.45 cm)=14.9 cm

   
   (c) [12 pts] A detector measures the intensity at the center of the viewing screen (y=0) to be 1.00 mW/m². What intensity does the detector measure at y=2.17 cm?
   
   

   y = Rtanq 2.17 = (120.)tan(q) q = 1.036o b 2 = pasinq l = p(10.0 mm)sin1.036o 0.620 mm = 0.292p I = Io é ë sin(b/2) b/2 ù û 2   = (1.00 mW/m2) é ë sin0.292p 0.292p ù û 2   = 0.750 mW/m2

FORMULAS

1/f=(n-1)(1/R1-1/R2) m=y¢/y 1/s+1/s¢=1/f dsinq = ml asinq = nl f = 2pdsinq/l b = 2pasinq/l y=Rtanq f=R/2 f = kDr Dy = Rl/d Dr=dsinq k=2p/l m=-s¢/s I=Iocos2(f/2) Dr=ml Dr=(m+1/2)l I=Io[sin(b/2)/(b/2)]2

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On 20 May 2016, 11:03.