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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

Let $I_1,I_2$ and $I_3$ be the image formed by
(ii) reflection from BEF and
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.

edited Jul 23, 2014
Thnk u so much !! Even i was confused in this concept!!