The image of an object formed by a mirror is real, inverted and its of magnification -1. If the image is at a distance of 40 cm from the mirror, where is the object placed ? Where would the image be if the object is moved 20 cm towards the mirror? State reason and also draw ray diagram for the new position of the object to justify your answer.

The image of an object formed by a mirror is real, inverted and its of

1.	The image of an object formed by a mirror is real, inverted and its of magnification -1. If the image is at a distance of 40 cm from the mirror, where is the object placed ? Where would the image be if the object is moved 20 cm towards the mirror? State reason and also draw ray diagram for the new position of the object to justify your answer.
Answer» Solution :MAGNIFICATION, m = -1, Image is real and inverted: Image distance = -40 cm `""....[because "image is real"]` `(-v)/(u) = m""rArr""(-(-40))/(u) = -1""rArr""u = (-(-40))/(-1)` `rArru = -40cm""therefore"""Object is placed at 40 cm in front of the mirror"`. When object distance is equal to the image distance and image is real, then the object is placed at C. If the object is MOVED 20cm towards themirror, then its newposition would be at the focus of the mirror. ltbr? According to the mirror formula, `(1)/(f)=(1)/(v)+(1)/(u)""rArr""(1)/(f)=(1)/(-40)+(1)/(-40)""rArr""(1)/(f)=(-1)/(40)-(1)/(40)` `rArr (1)/(f)=(-2)/(40)=(-1)/(20)""rArr""f = -20 cm` Second CASE : `f = -20 cm""u = -20 cm` `(1)/(v)+(1)/(u)=(1)/(f)""rArr""(1)/(v)+(1)/(-20)=(1)/(-20)` `rArr (1)/(v)-(1)/(20)=(-1)/(20)""rArr""(1)/(v)=(-1)/(20)+(1)/(20)` `rArr (1)/(v) = 0 rArr v = (1)/(0) prop""therefore` Thus, image will be formed at infinity.