question_answer

A concave mirror having the focal length 15 cm forms an image having twice of the linear dimensions of the object. If the image is virtual, then the position of the object will be

A) 7.5 cm             

B) 22.5 cm

C) 40 cm              

D) 30 cm

Correct Answer: A

Solution :

Image formation by concave mirror will be as follows: Given, \[m=\frac{image\,\,dis\tan ce}{object\,\,dis\tan ce}\,\,=\,\,\frac{\nu }{u}\,\,=\,\,2\] \[\Rightarrow \,\,\,\,\,\,\nu  =-1u\] Also \[\frac{1}{f}=\frac{1}{\nu }=\frac{1}{u}\] Where f is focal length, v is image distance and u is object distance. \[\Rightarrow \,\,\,\,\,\,-\frac{1}{15}=\frac{1}{u}-\frac{1}{2u}=\frac{1}{2u}\] \[\Rightarrow \,\,\,\,\,\,u=-\frac{15}{2}\,\,=\,\,-7.5\,cm\] Thus, the object should be placed at a distance 7.5 cm in front of the mirror.