Optics Question 954

Question: A plane mirror is made of glass slab $ ({\mu _{g}}=1.5)2.5cm $ 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:

Options:

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

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Answer:

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 $ BI _{1}=(5)({\mu _{g}})=(5)(1.5)=7.5cm $

Now, $ EI _{1}=(7.5+2.5)=10cm $

Hence, $ EI _{2}=10cm $ behind the mirror

$ BI _{2}=(10+2.5)=12.5cm $

$ \therefore $ $ BI _{3}=\frac{12.5}{{\mu _{g}}}=\frac{12.5}{1.5}=8.33cm $

The ray diagram is as follow:



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