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A plane mirror is made of glass slab $(\mu g=1.5)\;2.5\;cm$ thick and silvered on back. A point objected is placed 5 cm in front of unsilvered face of mirror. What will be the position of final image.

(a)8.33 cm from unsilvered face

(b)14.6 cm from unsilvered face

(c) 12 cm from unsilvered face  

(d) 5.67 cm from unsilvered face 

1 Answer

Let $I_1,I_2$ and $I_3$ be the image formed by
(i) refraction from ADC
(ii) reflection from BEF and
(iii) again refraction from ADC
Then $BI_1=5 \mu_g=5(1.5)=7.5 \;cm$
Now $\in I_1=(7.5 +2.5)=10 \;cm$
$\in I_2=10\;cm$ behind mirror.
Now $BI_2=(10+2.5) =12.5 \; cm$
$\therefore BI_3=\large\frac{12.5}{\mu g}=\frac{12.5}{1.5}$
$\qquad= 8.33 \;cm$
Hence a is the correct answer.


Thnk u so much !! Even i was confused in this concept!!
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