Physics, 19.03.2020 02:33 joThompson

His eyes are 1.68 m above the floor and the top of his head is 0.14 m higher. Find the height (in m) 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.

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Assume that the mirror and the person are both perpendicular to the ground. The bottom of that mirror should be at $0.84\; \rm m$ above the ground, while the top of the mirror should be at $1.75\; \rm m$ above the ground.

Explanation:

Refer to the diagrams attached (not to scale.) Let $\rm H$ and $\rm F$ denote the head and feet of this person, and let $\rm E$ denote his eyes.

Let the dashed vertical line in the middle of the diagram represent the location of the mirror. Let $\rm H^\prime$ denote the image of his head in the mirror, and let $\rm F^\prime$ be the image of his feet in the mirror.

Connect the eyes $\rm E$ and $\rm F^\prime$ , the image of the feet. Similarly, connect the eyes $\rm E$ and $\rm H^\prime$ , the image of the head.

For $\rm H^\prime$ , the image of the head to be visible in the mirror to $\rm E$ , the person's eyes, the line $\rm EH^\prime$ must go through the mirror. Similarly, for the image of the feet $\rm F^\prime$ to be visible in the image to $\rm E$ , the line $\rm FH^\prime$ must go through the mirror.

Let $\rm A$ be the intersection of $\rm EH^\prime$ and the mirror. Let $\rm B$ be the intersection of $\rm FH^\prime$ and the mirror. Based on that reasoning, $\rm AB$ should be the smallest mirror.

Both the person and the mirror are perpendicular to the ground. As a result, $\rm H^\prime F^\prime$ , the image of the person in the mirror, would also be perpendicular to the ground. Therefore, $\rm AB$ is parallel to $\rm H^\prime F^\prime$ , and the two triangles $\triangle \rm EAB$ and $\triangle \rm EH^\prime F^\prime$ would be similar to each other.

The distance between

the object and the mirror

should be the same as

the distance between the mirror and the image of the object.

Hence, the distance between the object and its image in the mirror should be twice the distance between the object and the mirror.

As a.result, the height of $\triangle \rm EH^\prime F^\prime$ on $\rm H^\prime F^\prime$ should be twice the height of $\triangle \rm EAB$ on $\rm AB$ . The ratio of similarity of these two triangles should be 2 :1 .

Therefore, the height of $\rm B$ (the bottom of the mirror) should be $\displaystyle \frac{1}{2}$ the vertical distance between the person's feet $\rm F$ and his eyes $\rm E$ . Hence, the height of the bottom of the mirror would be $\displaystyle \frac{1}{2} \times 1.68\; \rm m = 0.84\; \rm m$ .

Similarly, the vertical distance between $\rm A$ (the top of the mirror,) and $\rm E$ (his eyes) would be $\displaystyle \frac{1}{2}$ the vertical distance between his head $\rm H$ and his eyes $\rm E$ . Thus the height of $\rm A$ (the top of the mirror) would be above the height of the person's eyes by $\displaystyle \frac{1}{2} \times 0.14\; \rm m = 0.07\; \rm m$ . Since the person's eyes $\rm E$ are at a height of $\rm 1.68\; \rm m$ , the height of $\rm A$ (the top of the mirror) would be at $1.68\; \rm m + 0.07\; \rm m = 1.75\; \rm m$ relative to the ground.

His eyes are 1.68 m above the floor and the top of his head is 0.14 m higher. Find the height (in m)

Answer from: Quest

Answer: fn = wcos50 = 38.89(9.81)(cos50) = 245.23 n

Answer from: Quest

i guess it’s a and b

For which optical devices does d sometimes have a positive value

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His eyes are 1.68 m above the floor and the top of his head is 0.14 m higher. Find the height (in m)...

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