What is the width of a wall seen from an opposing mirror?

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SUMMARY

The discussion focuses on a geometry problem involving two parallel walls in a church choir loft, spaced 4.98 meters apart. The organist, positioned 0.967 meters from the south wall, uses a 0.855-meter-wide flat mirror mounted on that wall to view the choir against the north wall. The key to solving the problem lies in applying the Law of Reflection rather than Snell's Law, as the angles of incidence and reflection are equal, allowing for the calculation of the visible width of the north wall.

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Aneadra
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Homework Statement



In a church choir loft, two parallel walls are
4.98 m apart. The singers stand against the
north wall. The organist faces the south wall,
sitting 0.967 m away from it. So that she can
see the choir, a flat mirror 0.855 m wide is
mounted in the south wall, straight in front of
the organist.
What width of the north wall can she see?
Answer in units of m.

Homework Equations


I'm not exactly sure, but I think snells law is incorporated somehow.

The Attempt at a Solution


I drew a diagram showing how the rays would bounce off the north wall into the southwall and I do see where some triangles are formed, other than that I'm kinda lost...
 
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Why not trace some rays from the head of the organist to the edges of the mirror and across to the opposite wall?
 
Aneadra said:
I'm not exactly sure, but I think snells law is incorporated somehow.

Mmmm, not quite. How about the Law of Reflection?
 

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