Consider two waves defined by the wave functions y1(x,t)=0.50msin(2π3.00mx+2π4.00st) and y2(x,t)=0.50msin(2π6.00mx−2π4.00st). Wh
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This question is incomplete, the complete question is;
A woman is 160cm tall. What is the minimum vertical length of a mirror in which she can see her entire body while standing upright.
Hint: Consider the ray diagram below, of the rays that enable her to see her feet and the top of her head.
Use what you know about the law of reflection, together with a bit geometry.
The missing Image is uploaded along this answer below.
Answer:
the minimum vertical length of a mirror in which she can see her entire body while standing upright is 80 cm
Explanation:
Given the data in the question and illustrated in the image below,
From image 2;
The distance from the woman's eyes to the top of her head is represented as b and a represent the distance from her eyes to her feet.
Therefore, since her height is 160 cm
a + b = 160 ------ let this be equation 1
i.e AD + DG = 160 cm
Now, from the same image 2, we will notice that triangle ABC and tringle CBD are similar, so
∠ABC = ∠CBD
AC = CD
since AD = a and AC + CD = A
AC = CD = a/2
Also, triangle DEF and FEG are si,ilar
∠DEF = ∠FEG
so
DF = FG
since DG = b and DF + FG = b
DF = b/2
so the minimum vertical length of a mirror in which the woman can see her entire body while standing upright will be;
⇒ a/2 + b/2
⇒ a + b / 2
from equation 1, a + b = 160
so
⇒ a + b / 2 = 160 / 2 = 80
Therefore, the minimum vertical length of a mirror in which she can see her entire body while standing upright is 80 cm
