question_answer

A 2.0 cm tall object is placed 15 cm in front of a concave mirror of focal length 10 cm. What is the size and nature of the image

A) 4 cm, real         

B)        4 cm, virtual

C)        1.0 cm, real                  

D)        None of these

Correct Answer: A

Solution :

According to New Cartesian sign convention, \[\operatorname{Object} distance u = - 15 cm\] Focal length of a concave lens, \[\operatorname{f}= -10 cm\] Height of the object \[{{h}_{0}}=2.0 cm\] According to mirror formula, \[\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\] \[\frac{1}{v}+\frac{1}{f}=\frac{1}{u}=\frac{1}{-10}-\frac{1}{-15}\Rightarrow \,\,v=-\,30cm\] This image is formed 30 cm from the mirror on the same side of the object. It is real image. Magnification of the mirror, \[\operatorname{m} =\frac{-\nu }{u}=\frac{{{h}_{1}}}{{{h}_{0}}}\] \[\Rightarrow \,\,\,\,\frac{-(-30)}{-15}=\frac{{{h}_{1}}}{2}\Rightarrow \,\,{{h}_{1}}=-\,4\,cm\] Negative sign shows that image is inverted. The image is real, inverted, of size 4 cm at a distance 30 cm in front of the mirror.