Minimum length of the mirror required for a man to see the wall behind him?

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SUMMARY

The minimum length of a mirror required for a man of height 2h to see the wall behind him in a room of length 6h and wall height 4h is (4/3)h. The solution involves geometric principles, specifically the properties of similar triangles. By positioning the viewer at the midpoint of the room and analyzing the reflection path, the necessary calculations yield the mirror length. The derived formula confirms that the mirror must extend to (4/3)h to ensure full visibility of the wall.

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Lakshya
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A man of height 2h stands in the middle of a room of length 6h. The height of a wall is 4h. What is the minimum length of a mirror to be placed on the wall in front of the man so that he can see the entire wall behind him?
 
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Lakshya said:
A man of height 2h stands in the middle of a room of length 6h. The height of a wall is 4h. What is the minimum length of a mirror to be placed on the wall in front of the man so that he can see the entire wall behind him?

What is your approach to the problem? What conditions have to be met?
 
LowlyPion said:
What is your approach to the problem? What conditions have to be met?
The question is clear as it is. By the way, the answer is (4/3)h. I want the solution.
 
Lakshya said:
The question is clear as it is. By the way, the answer is (4/3)h. I want the solution.

Forum rules are you will have to find the solution yourself.

I was offering to help. If you need help, you can explain what you've done.

As to the answer, I already had that.
 
LowlyPion said:
Forum rules are you will have to find the solution yourself.

I was offering to help. If you need help, you can explain what you've done.

As to the answer, I already had that.

Yes, I need help. Tell me how to solve this problem. I don't know even a bit how to solve it.
 
Lakshya said:
Yes, I need help. Tell me how to solve this problem. I don't know even a bit how to solve it.

I would suggest that you draw a diagram with the viewer in the middle of the room at 3H. Divide the room vertically in half at 2H (to solve the top half which we can then double) with the eye level at 2H high now, the ray trace should go from that mid-point to one wall above the 2H level that reflects back to the top of the wall behind the viewer another 6H away. That total distance describes a triangle (if you unfolded the trip to and from that is 9H long and 2H high.

Since it is a similar triangle, the distance at the wall (the top height of the mirror) is to 3H as 2H is to 9H

X/3H = 2H/9H
X = 2H/3

Your final answer is 4H/3
 

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