Question

A light bulb burns in front of the center of a $40$cm wide plane mirror that is hung vertically on a wall. A man walks in front of the mirror along a line that is parallel to the mirror and twice as far from it as the bulb. The greatest distance he can walk and still see the image of the bulb.

• A. $20$cm
• B. $40$cm
• C. $60$cm
• D. $120$cm

Answer 1

The correct option is D $120$cm

Given: A light bulb burns in front of the center of a 40cm wide plane mirror that is hung vertically on a wall. A man walks in front of the mirror along a line that is parallel to the mirror and twice as far from it as the bulb.

To find the greatest distance he can walk and still see the image of the bulb

Solution:

According to the given criteria,

the ray diagram is as shown above,

Width of plane mirror, AB = 40cm

Let the distance between bulb and mirror be $L$

Hence the distance between man and mirro is $2L$

$HI=AB=40cm$

and, In $\Delta ASC$,

AD is perpendicular bisector

Therefore, $DS=CD=\frac{40}{2}=20cm$

As $AH=2AD⟹GH=2CD=\frac{2\times 40}{2}=40cm$

Similarly,

$IJ=40cm$

So, $GJ=GH+HI+IJ=40+40+40=120cm$

is the greatest distance he can walk and still see the image of the bulb.