question_answer

A plane mirror is made of glass slab \[({{\mu }_{g}}=1.5)\,2.5\,\text{cm}\] thick and silvered on the back. A point object is placed 5 cm in front of the unsilvered face of the mirror. What will be the position of final image:

A) 12 cm from front face                        

B) 14.6 cm from front face

C) 5.67 cm from front face                      

D) 8.33 cm from front face

Correct Answer: D

Solution :

[d] Let, \[{{I}_{1}},{{I}_{2}}\] and \[{{I}_{3}}\] be the images formed by:

(i) refraction from ABC

(ii) reflection from DEF and

(iii) again refraction from ABC

Then \[B{{I}_{1}}=(5)({{\mu }_{g}})=(5)(1.5)=7.5\,cm\]

Now,    \[E{{I}_{1}}=(7.5+2.5)=10\,cm\]

Hence, \[E{{I}_{2}}=10\,cm\] behind the mirror

\[B{{I}_{2}}=(10+2.5)=12.5\,\,cm\]

\[\therefore \]    \[B{{I}_{3}}=\frac{12.5}{{{\mu }_{g}}}=\frac{12.5}{1.5}=8.33\,cm\]

The ray diagram is as follow: