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A light ray is incident on a horizontal plane mirror at an angle of 45^(@). At what angle should a second plane mirror be placed in order that the refelcted ray finally be reflected horizontally form the second mirror, as shown in figure |
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Answer» `theta=30^(@)` So, `/_DCB/_CBQ=180^(@)` `/_DCN+/_NCB+/_CBQ=180^(@)` `alpha+alpha+45^(@)=180^(@)` `alpha=67.5^(@)` But `/_NCS=90^(@)` So, SECOND mirror is at an ANGLE of `22.5^(@)` with horizontal . |
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