1. By what length is a light ray displaced after passing through a 5.7 cm thick
ID: 1285814 • Letter: 1
Question
1. By what length is a light ray displaced after passing through a 5.7 cm thick sheet of material (n=1.45) with an incident angle of theta = 31.5.
2.) Suppose a man stands in front of a mirror as shown in the figure below. His eyes are 1.65 m above the floor, and the top of his head is 0.130 m higher. Find the height above the floor of the top and bottom of the smallest mirror in which he can see both the top of his head and his feet. How is this distance related to the mans height?
Figure caption: A fulllength mirror is one in which you can see all of yourself. It need not be as big as you, and its size is independent of your distance from it.
(in m) Floor to bottom of mirror =
(in m) Floor to top of mirror =
Explanation / Answer
1)
The lateral displacement = t * sin(i -r)/(cos(r))
where i = angle of incidence
r=angle of refraction
t=thickness of the slab
we know
n1*sin(i)=n2*sin(r)
n1=1
n2=1.45
sin(31.5)=1.45*sin(r)
sin(r)=0.36
r=21.12o
lateral displacement = 5.7 * sin(31.5-21.12)/cos(21.12)
=1.1cm
b)
Let height above ther person's eye be h1 and height below his eyes be h2
The length of the mirror required above his eye to see till his head =(h1)/2
The length of mirror required to see till his toe below his eye=(h2)/2
[A short proof:
The angle of incidence = angle of reflection for any part
The perpendicular dropped from mirror to man is common
The angle is right angle
The triangles formed are congruent and hence the lenght of mirror required is half of any part]
The bottom of the mirror =h2/2 =1.65/2 =0.825
The top of the mirror = h2 + h1/2 =1.65+0.130/2= 1.65+0.065 =1.715m
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