Question

Three transparent slabs are arranged one over another as shown in the figure. The refractive indices and the thickness are as given in the figure. An object is seen through this combination with nearly perpendicular light. Find the equivalent refractive index of the system which will allow the image to be formed at the same place.

• A. $\frac{6}{5}$
• B. $\frac{4}{3}$
• C. $\frac{5}{4}$
• D. $\frac{3}{2}$

Answer 1

The correct option is B $\frac{4}{3}$

In the given situation, the total shift will be given by
$\Delta t=(1-\frac{1}{{\mu }_1})\left(3\right)+(1-\frac{1}{{\mu }_2})\left(4\right)+(1-\frac{1}{{\mu }_3})\left(5\right)\dots \left(1\right)$
If $\mu$ is the equivalent refractive index of the combination of slabs and the image is formed at same place,
$\Delta t=(1-\frac{1}{\mu })(3+4+5)\dots \left(2\right)$

From equation (1) and (2),
$(1-\frac{1}{\mu })(3+4+5)=(1-\frac{1}{{\mu }_1})\left(3\right)+(1-\frac{1}{{\mu }_2})\left(4\right)+(1-\frac{1}{{\mu }_3})\left(5\right)$
$\Rightarrow -\frac{1}{\mu }(3+4+5)=-\frac{3}{{\mu }_1}-\frac{4}{{\mu }_2}-\frac{5}{{\mu }_3}$
$\Rightarrow \mu =\frac{(3+4+5)}{\frac{3}{\frac{3}{2}}+\frac{4}{\frac{4}{3}}+\frac{5}{\frac{5}{4}}}$
$=\frac{12}{9}=\frac{4}{3}$

Why this question? NOTE: In case of multiple slabs, find the apparent shift due to each slab and add them to get the total shift.