Determine the smallest size of a plane mirror which will enable a man of 1.8 metre height to see his full image.
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Let MN be the mirror placed in the vertical position XY in front of man AB with eyes at E.
draw AX perpendicular to XY and produce it to c, so that AX=XC, then C is the image of A.
Similarly , draw BY perpendicular to XY and produce it to D, so that BY=YD . Then D is the image of B.
CD is the image of AB in the mirror as the image of any object in plane mirror is virtual of same size , erect and as far behind as the object is in font of it.
To find the rays of light by which the image is seen by the eye, join E to c and D cutting XY at M and N.
Join A to M and B to N, so that the ray Am starting from top of man appears to reach the eye E from C after refection at M.
Similarly, The ray Bn starting from the foot of the man reaches the eye E after reflection from the mirror at N. so the required sixe of mirror is MN.
In Triangle, the point X is the midpoint of AC and XM is parallel To AE, so M is the midpoint of CE, similarly, N is the midpoint of DE.
In triangle, CED, M and N are midpoints of sides CE and DE . and MN parllel to Cd
So, MN=1/2CD
=1/2AB
hence, MN=1/2x 1.8 =0.9 m
So that MN or the smallest size of required mirror is 0.9m
draw AX perpendicular to XY and produce it to c, so that AX=XC, then C is the image of A.
Similarly , draw BY perpendicular to XY and produce it to D, so that BY=YD . Then D is the image of B.
CD is the image of AB in the mirror as the image of any object in plane mirror is virtual of same size , erect and as far behind as the object is in font of it.
To find the rays of light by which the image is seen by the eye, join E to c and D cutting XY at M and N.
Join A to M and B to N, so that the ray Am starting from top of man appears to reach the eye E from C after refection at M.
Similarly, The ray Bn starting from the foot of the man reaches the eye E after reflection from the mirror at N. so the required sixe of mirror is MN.
In Triangle, the point X is the midpoint of AC and XM is parallel To AE, so M is the midpoint of CE, similarly, N is the midpoint of DE.
In triangle, CED, M and N are midpoints of sides CE and DE . and MN parllel to Cd
So, MN=1/2CD
=1/2AB
hence, MN=1/2x 1.8 =0.9 m
So that MN or the smallest size of required mirror is 0.9m
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