QUIZ 23 (TEST 4)  AND (TEST 3)

 Index of refraction lab: 

PROBLEMS: 3,  4, 7, 8, 9, 10, 11, 12, 15, 16, 20, 24, 25, 30, 31, 32, 38, 39, 43, 44, 45, 46, 47, 48, 49, 52

TURN IN:  4, 8, 10, 12 (EC), 16, 20(EC), 24, 30, 32, 38, 44(EC), 46(EC), 48, 52
4. You need the angle theta that the ray between the mirror and the nearest point on the floor that can be seen reflected in the mirror.  Then you'd use tan (theta) = 0.43/x. 

You can find theta from the horizontal distance 2.20 m and the vertical length of the mirror. The correct ratio of these lengths gives you tan (theta)  which allows you to compute x.   
8. Use equation 23-2. You can get f from  given R. Set di = infinity.  Solve for do. 
10. An upright image occurs in the case of 
(i) a convex mirror or 
(ii) a  concave mirror when the do < f.   

However, only in case (ii) can the magnification be greater than 1.  Thus use: 

1/di   + 1/do  = 1/f  

and

m = - di/do 


to simultaneously solve for f and di. Note that m = + 3 and     do = 1.4 m.  Thus  di is negative.

16. This seems to be a convex mirror since  magnifications m < 1. di = - 0.18 m and do = infinity. Use 1/di   + 1/do  = 1/f  to find f. From this, find R.
24. v = c/n.
30.

(a) Use equation 23-5 to find the angle with the normal when the ray is in the glass assuming the first angle is 43.5 degrees. 

n1*sin(theta1) = n2*sin(theta2) where 1 denotes the ray's  incidence and reflection medium and 2 denotes it's transmission medium. 

Solve for the second angle. See figure 23-22. 

(b) Use equation 23-5 to find the angle with the normal when the ray is in the water assuming the first angle i has the value of the second angle 2 you solved for in part (a). . 

 ni*sin(theta_i) = nf*sin(theta_f) where i denotes the ray's  incidence and reflection medium and f denotes it's transmission medium. Solve for the final  angle f. 

(c)  Use equation 23-5 to find the angle with the normal when the ray passes directly from air into water assuming the first angle is 43.5 degrees.  See figure 23-19.

n1*sin(theta1) = n2*sin(theta2) where 1 denotes the ray's  incidence and reflection medium and 2 denotes it's transmission medium. 
32. You really need a picture for this but here goes: You can easily find the second angle, theta2, shown using Snell's law: n1*sin(theta1) = n2*sin(theta2) where 1 denotes the ray's  incidence and reflection medium and 2 denotes it's transmission medium. 

With theta 2,  you must geometrically find the angle with the opposite face's normal theta3. Here's a suggestion: 
Look slightly above the center of the glass triangle. You will see and internal triangle formed from the red  and the two dashed lines inside the glass.  The internal angles of this small triangle must sum to 180 degrees.  Now you already know theta2 from Snell's Law. It can be shown geometrically that the small triangle's  lowest corner has internal angle = 2*60 = 120 degrees.

The above result depends on (A) the angle between two lines that are  perpendicular to opposite equilateral triangle faces and  (B) the  equilateral triangle's equal internal angles of 60 degrees. 

Thus, we have:

theta2 + 120 + theta3 = 180. Solve for theta3 and substitute it into Snell's law  as theta_i:

ni*sin(theta_i) = nf*sin(theta_f) . 

Solve for theta_f. 
38. Zoom in on figure 23-24. From the critical angle you want to find other quantities geometrically.  Let x be the horizontal beam displacement between the exit point of a vertically fired  beam and  actual exit point. Clearly, 
x/0.62 = tan(theta_c) where theta_c is the critical angle for a water-air interface.  
48. f = -1/(5.5) m. Use 1/di + 1/do = 1/f. Find the image distance given  do.  Then find magnification m to compute  hf from hi = 0.0040 m. Is the image virtual or real? Assuming the object is upright, is the image upright or inverted? 
52. Let m = -1.  That means -1 = -di/do  or  1 = di/do .  That is,  di   =  do. If the image is real, then do > f. Thus, di > f and di = do.  Plug in symbols and solve for do in terms of f:  1/di + 1/do =  = 1/do + 1/do = 1/f . What is  do ? Is  do > f ? 
more hints later.