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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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by srinathg

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by srinathg

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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.