Question

Two slabs of same thickness 1 cm each are placed as shown.

Find the refractive index of a slab of thickness 3 cm which will produce the same normal shift in position of object (O), as produced by combination of slabs.

• A. 1.65
• B. 1.38
• C. 2.15
• D. 1.25

Answer 1

The correct option is B 1.38

Let ${\mu }_{eq}$ be the R.I of slab of thickness $t=3cm$
$\Rightarrow S_{net}=S_1+S_2$
or $t(1-\frac{1}{{\mu }_{eq}})=t_1(1-\frac{1}{{\mu }_1})+t_2(1-\frac{1}{{\mu }_2})$
$3(1-\frac{1}{{\mu }_{eq}})=1(1-\frac{2}{3})+1(1-\frac{1}{2})$
$\Rightarrow 3(1-\frac{1}{{\mu }_{eq}})=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}$
or, $1-\frac{1}{{\mu }_{eq}}=\frac{5}{18}$
or, $\frac{1}{{\mu }_{eq}}=1-\frac{5}{18}=\frac{13}{18}$
$\therefore {\mu }_{eq}=\frac{18}{13}=1.38$

Why this question? Caution: The formula ${\mu }_{eq}=\frac{2{\mu }_1{\mu }_2}{{\mu }_1{\mu }_2}$ is only applicable when sum of thickness of two slabs is equal to thickness of equivalent slab.