(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

(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

(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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