QUIZ 24,  CH.24 (REAL TEST 4)
 QUIZ 24  
SELECTED DISCUSSIONS TO exercises/problems BELOW.
YOU ARE RESPONSIBLE FOR ALL  PROBLEMS EVEN THE ONES WITHOUT DISCUSSIONS, WHICH ARE DESIGNED TO HELP YOU DO RELATED EXERCISES. WE WILL REVIEW THEM IN-CLASS. ALSO READ ALL EXAMPLES; THE EXERCISES ARE  SIMILAR.
CH.  24
CH. 24  MULTIPLE CHOICE TBA  -
TURN IN: CH. 24---5, 6, 10, 12, 13, 14, 29, 30, 33, 34, 50, 52, 53
* DISCUSSIONS PROVIDED. 
5. Use 1/f = 1/di  + 1/do and f = R/2 > 0. Find di and m = -di/do,  from which you may get final height hf = m*hi; note hf is negative if initial height hi is positive if the image is inverted and real. That's because di > 0 and m = -di/do < 0. That happens when do > f. When do < f, the image is virtual and upright and di <0, making m = -di/do > 0. See # 12 below. 
6. Let f = -|R|/2. Use 1/f = 1/di  + 1/do. Find di and m = -di/do,  from which you may get final height hf = m*hi; note hf is positive if initial height hi is positive since the image is upright and virtual . Note that di < 0 in all cases. Thus m > 0.
10.  (a) Let f = -|R|/2. Use 1/f = 1/di  + 1/do. Find di and m = -di/do,  from which you may get final height hf = m*hi; note hf is positive if initial height hi is positive since the image is upright and virtual . Note that di < 0 in all cases. Thus m > 0. 
(b) Use 1/f = 1/di  + 1/do. Find di and m = -di/do.
12. (a) Use SYMBOLS: Use 1/f = 1/di  + 1/do and f  > 0. Find di and m = -di/do if do > f. Show di > 0   from which you get final height hf = m*hi; note hf is negative if initial height hi is positive since the image is inverted and real.  That's because  m = -di/do < 0.
(b) Use SYMBOLS: Use 1/f = 1/di  + 1/do and f  > 0. Find di and m = -di/do if do < f. Show di < 0   from which you get final height hf = m*hi; note hf is positive if initial height hi is positive since the image is upright and virtual. That's because  m = -di/do is now positive!
13. Use 1/f = 1/di  + 1/do and f = R/2 > 0. Find di and m = -di/do,  from which you may get final height hf = m*hi; note hf is negative if initial height hi is positive if  the image is inverted and real . That happens when do > f. When do < f, the image is virtual and upright and di <0, making m = -di/do > 0. See # 12 above. 
(a) Find m.
(b) Find di.
(c) See distributed class notes where we drew strategic rays to map out, through ray tracing,  the location of the image given an object,  its  distance from mirror do and  height ho. In this case you are given do but no number for ho. Still you can draw an arrow at the location of the object, distance do from mirror surface at axis. The arrow tail is on the axis line of mirror and the tip is vertically above that point some finite distance defining arrow's height ho. 
14. Use 1/f = 1/di  + 1/do and f = R/2 > 0. Find di and m = -di/do,  from which you may get final height hf = m*hi; note hf is negative if initial height hi is positive if  the image is inverted and real . That happens when do > f. When do < f, the image is virtual and upright and di <0, making m = -di/do > 0. See # 12 above. 
(a)  See distributed class notes where we drew strategic rays to map out, through ray tracing,  the location of the image given an object,  its  distance from mirror do and  height ho. In this case you are given do and  number for ho. Still you can draw an arrow at the location of the object, distance do from mirror surface at axis. The arrow tail is on the axis line of mirror and the tip is vertically above that point some finite distance defining arrow's height ho..
(b) Find di, m and hf given ho > 0. Is di > 0 or di < 0? Is the image real (inverted) or real (upright); in other words is m < 0 or m >0, where m = - di/do?
29. Use 1/f = 1/di  + 1/do and f  > 0. Find di and m = -di/do,  from which you may get final height hf = m*hi; note hf is negative if initial height hi is positive if  the image is inverted and real . That happens when do > f. When do < f, the image is virtual and upright and di <0, making m = -di/do > 0.

The object is to the left of lens since the virtual image is stated to be upright like the object; see
fig. 24. 37 (e).  You have two unknowns, di and do,  and two equations:
Find  (i)1/f = 1/di  + 1/do , where  f  > 0 and   (ii) hf = -(di/do)*hi = m*hi, where m = -di/do.
(ii) may be re-written and  di = -(1.30 cm /0.400 cm)*do or
(ii) di = -3.25do .

Substitute  (ii) into (i):
1/f = -1/(3.25do)  + 1/do
Solve for do and di. Note: do > 0 and di < 0.
30. Use 1/f = 1/di  + 1/do and f  > 0. Find di and m = -di/do,  from which you may get final height hf = m*hi; note hf is negative if initial height hi is positive if the  image is inverted and real . That happens when do > f. When do < f, the image is virtual and upright and di <0, making m = -di/do > 0.

The object is to the right of lens since the real image is stated to be inverted (unlike the object) and on the left of the lens; see figs. 24. 37 (a), (b), (c); you might have to rotate the figures to fit this problem such that  object is to right of lens.

You have two unknowns, di and do,  and two equations:
Find  (i)1/f = 1/di  + 1/do , where  f  > 0 and   (ii) hf = -(di/do)*hi = m*hi, where m = -di/do.
(ii) may be re-written and  di = -(-4.50 cm /3.20 cm)*do or (ii) di = +1.406do .

Substitute  (ii) into (i):
1/f = 1/(1.406do)  + 1/do
Solve for do and di. Note: do > 0 and di > 0.
MORE DISCUSSIONS TO FOLLOW----COME TO CLASS AND CHECK THIS SITE AGAIN.