Question

5. The focal length of a concave mirror is 10 cm.
Where should an object be kept in front of the
mirror so that its formed image be 5 times of the
object, while image be : (i) real (ii) virtual ? Also
find the position of image.
[Ans. Keeping object at a distance of 12 cm and
image is formed 60 cm, object will keep at a
distance of 8 cm in front of mirror and image will
formed at 40 cm]


please answer this ques in physics numerical method​

Answer 1

m= f/f-u,. here f= -10cm , u=?, here two condition is possible m= -5 (for real image), m= 5 (for virtual image),. 1. -5 = -10/-10-u,. 50+5u= -10 , 5u= -10-50,. 5u= -60,. u= -12 cm. using another magnification formula for calculate image distance is m= -v/u,. -5= -v/-12 , v= -60 cm ,. 2. m= 5 ,. 5= -10/-10-u,. -50-5u= -10,. -5u= -10+50, -5u =40 , u= -8 cm,. using m= -v/u,. 5= -v/-8 ,. v= +40 cm