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: |
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