A ray of light incident at 60degree on the Middle pair of plane mirrors arranged at 60degree to each other.Calculate the angle of incidence in the second plane mirror
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Two plane mirrors AB and BC are arranged at 60 deg. to each other. so angle ABC = 60 deg.
AB and BC are the reflecting surfaces of the mirror..
A ray CD is incident on the mirror AB , meeting AB at D. Let the normal to AB at D be DE. So the angle of incidence CDE is 60 deg.
The ray is reflected along the line DF , where the angle EDF = angle of reflection = 60 deg.
Let DF meet the mirror BC at G.
Now look at the triangle DBG. angle GDB = 90 - 60 = 30 deg.
angle ABC = 60 deg. = angle DBG
Hence, the angle BGD = 180 - 30 - 60 = 90 deg.
So the reflected ray from one mirror is incident on the other mirror at 90 deg. Hence, it is reflected back normally. and it falls on the same point on the first mirror. So the ray retraces its path.
AB and BC are the reflecting surfaces of the mirror..
A ray CD is incident on the mirror AB , meeting AB at D. Let the normal to AB at D be DE. So the angle of incidence CDE is 60 deg.
The ray is reflected along the line DF , where the angle EDF = angle of reflection = 60 deg.
Let DF meet the mirror BC at G.
Now look at the triangle DBG. angle GDB = 90 - 60 = 30 deg.
angle ABC = 60 deg. = angle DBG
Hence, the angle BGD = 180 - 30 - 60 = 90 deg.
So the reflected ray from one mirror is incident on the other mirror at 90 deg. Hence, it is reflected back normally. and it falls on the same point on the first mirror. So the ray retraces its path.
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