question_answer

A mirror is broken into two parts and these parts are separated by a distance of 1 cm as shown in figure. The focal length of the mirror is 10 cm. The height of the images (in cm), formed by the two parts of mirror when the object is midway between the two principal axes are

A) 1, 1                  

B)        1, 2                  

C) 2, 2                  

D)        0, 5, 1

Correct Answer: A

Solution :

We know, \[\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\] Here, \[f=-10\,cm,\,u=-15\,cm\] \[\therefore \]      \[\frac{1}{-10}=\frac{1}{-15}+\frac{1}{v}\] or,\[\frac{1}{v}=\frac{1}{15}-\frac{1}{10}=-\frac{1}{30}\] or, \[v=-30\,cm\] We know, \[m=-\frac{v}{u}=-\frac{-30}{-15}=-2\] Distance between broken mirrors is 1 cm. Hence, two images will be formed at a distance of 30 cm from the mirror. \[{{h}_{i1}}=m{{h}_{o1}}=(-2)\times (-0.5)=1\,cm\] Image \[({{l}_{1}})\] due to upper part of the mirror will be formed at 1 cm above its optic axis. Similarly, image\[({{l}_{2}})\]due to lower part of the mirror will be formed at 1 cm below its optics axis.