Reflection and Mirrors Review
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Questions #33#42
Part C: Calculations and Explanations
33. Distinguish between diffuse and regular (specular) reflection in terms of both cause and result.
Answer:
Specular (or regular) reflection occurs when light reflects off a microscopically smooth surface. Light rays which are incident within a beam will reflect and remain in the beam.
Diffuse reflection occurs when light reflects off a microscopically rough surface. Light rays which are incident in a beam will diffuse (or scatter) and reflect in a variety of directions.
In each case the law of reflection holds; in diffuse reflection, the normals for each individual ray are not oriented in the same direction.

34. Write the formulae which show the relationship between image distance, object distance and focal length, and the magnification of an object/image.
Answer:
Please know this one:
where s and h stand for object distance and object height,
and s' and h' stand for image distance and image height,
and f stands for focal length,
and M stands for magnification.

35. Fill in the blanks below for the problemsolving rules:
Focal lengths, object and image distances are ________ if they are on the reflective side of the mirror. Object and image heights are positive if they are ________ the principle axis.
Answer:
Focal lengths, object and image distances are positive if they are on the reflective side of the mirror. Object and image heights are positive if they are below the principle axis.
The d_{o} and d_{i} values (s and s') and the h_{o} and h_{i} values (h and h') can have positive and negative values associated with them. In physics, a + or  in front of a quantity is always descriptive of some physical feature. In ray optics, a + distance (d_{o}, d_{i} or f) means in front of the mirror as the object; a negative distance means behind the mirror. A + height means above the principal axis and a  height means below the principal axis.

36. What is the focal length of a mirror which gives an image 3.00 meters away when an object is placed 150. cm from it?
Answer: 100. cm or 1.00 m
Use the mirror equation:
1/d_{i} + 1/d_{o} = 1/f
where d_{i} = +300. cm and d_{o} = +150. cm. (Note: since d_{i} > d_{o}, it must be a concave mirror and the image must be real.)
Substitute and solve for f.
1/(150. cm) + 1/(300. cm) = 1/f
1/f = 3.00/(300. cm) or 0.0100/cm
f = 100. cm = 1.00 m

37. Where is the image located when an object is placed 60.0 cm from a mirror which has a focal length of 20.0 cm?
Answer: 30.0 cm
Use the mirror equation:
1/d_{i} + 1/d_{o} = 1/f
where d_{0} = +60.0 cm and f = +20.0 cm.
Substitute and solve for d_{i}.
1/(60.0 cm) + 1/d_{i }= 1/(20.0 cm)
1/d_{i} = 1/(20.0 cm)  1/(60.0 cm) = 2/(60.0 cm)
d_{i} = 30.0 cm

38. How far would an object need to be placed from a mirror of focal length 10.0 cm if it is to produce an image which is 20.0 cm BEHIND the mirror?
Answer: 6.67 cm
Use the mirror equation:
1/d_{i} + 1/d_{o} = 1/f
where d_{i} = 20.0 cm and f = +10.0cm.
Substitute and solve for d_{o}.
1/d_{o }+ 1/(20.0 cm) _{ }= 1/(10.0 cm)
1/d_{o} = 1/(10.0 cm)  1/(20.0 cm) = 3.00/(20.0 cm) = 0.150/cm
d_{o} = 6.67 cm

39. A mirror with a focal length of 100. cm is used to form an image. An object is placed 50.0 cm in front of the mirror.
a. Where is the image located?
b. What type of mirror is being used in this problem?
Answer: 33.3 cm; convex mirror
Use the mirror equation:
1/d_{i} + 1/d_{o} = 1/f
where d_{0} = +50.0 cm and f = 100. cm. (Note: this is a convex mirror since the focal length is negative.)
Substitute and solve for d_{i}.
1/(50.0 cm) + 1/d_{i }= 1/(100. cm)
1/d_{i} = 1/(100. cm)  1/(50.0 cm) = 3.00/(100. cm) or 0.03/cm
d_{i} = 33.3 cm

40. An object is placed 20 cm from a mirror of focal length 10.0 cm. The object is 5.0 cm tall. Where is the image located? How tall is the image?
Answer: d_{i} = 20.0 cm; h_{i} = 5.0 cm
Use the mirror equation:
1/d_{i} + 1/d_{o} = 1/f
where d_{0} = +20.0 cm and f = 10.0 cm.
Substitute and solve for d_{i}.
1/(20.0 cm) + 1/d_{i }= 1/(10.0 cm)
1/d_{i} = 1/(10.0 cm)  1/(20.0 cm) = 1/(20.0 cm) or 0.0500/cm
d_{i} = 20.0cm
Use the magnification ratio to solve for h_{i}:
h_{i}/h_{o} = d_{i}/d_{o}
where h_{o} = 5.0 cm.
h_{i }/(5.0 cm) =  (20.0 cm)/(20.0 cm)
h_{i} =  [(20.0 cm)/(20.0 cm)]•(5.0 cm)
h_{i} = 5.0 cm

41. A 20.0 cm tall object produces a 40.0 cm tall real image when placed in front of a curved mirror with a focal length of 50.0 cm. Determine the place that the object must be located to produce this image.
Answer: 75 ,.0cm
Given: h_{o} = 20.0 cm, h_{i} = 40.0 cm, f = 50.0 cm (Note: h_{i} is negative since the image is real.)
Strategy: use the object and image height values with the magnification ratio in order to generate an expression for d_{i} in terms of d_{o}. Then substitute the expression into the mirror equation to solve for the object distance.
h_{i}/h_{o} = d_{i}/d_{o}
(40.0 cm)/(20.0 cm) = d_{i}/d_{o}
2.00 = = d_{i}/d_{o}
d_{i }= 2.00*d_{o}
Now substitute this expression into the mirror equation.
1/d_{o}+ 1/(2.00*d_{o}) = 1/(50.0 cm)
2/(2.00*d_{o}) + 1/(2d_{o}) = 1/(50.0 cm)
3/(2.00*d_{o}) = 1/(50.0 cm)
2.00*d_{o}= 150. cm
d_{o} = 75.0 cm

42. A mirror is used to produce a virtual image that is 5.00 times bigger than the object. If the focal length of the mirror is 100. cm, where will the image be located?
Answer: 400. cm
Given: M = +5.00 , f = 100. cm (Note: M is positive since the image is virtual.)
Strategy: use the magnification value with the magnification ratio in order to generate an expression for d_{o} in terms of d_{i}. Then substitute the expression into the mirror equation to solve for the object distance.
M = h_{i}/h_{o} = d_{i}/d_{o}
+5.00 = d_{i}/d_{o}
d_{o }= = d_{i}/5.00
Now substitute this expression into the mirror equation.
1/(d_{i}/5.00)+ 1/(d_{i}) = 1/(100. cm)
5.00/(d_{i}) + 1/(d_{i}) = 1/(100. cm)
4.00/(d_{i}) = 1/(100. cm)
d_{i }= 400. cm

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