Question

A slab of transparent material is made as shown in the figure. The thickness of $C$ is twice the thickness of $B$. If the number of waves in $A$ is equal to the number of waves in combination of $B$ and $C$, then the refractive index of $B$ is

• A. $1.8$
• B. $1.6$
• C. $1.4$
• D. $1.33$

Answer 1

The correct option is A $1.8$

Let the thickness of $B=t$
$\therefore$ thickness of $C=2t$
thus, thickness of $A=t+2t=3t$
if ${\lambda }_0$ is wavelength in air, then as number of waves in $A$ is equal to number of waves in $B+$ number of waves in $C$
$\Rightarrow \frac{3t}{\frac{{\lambda }_0}{{\mu }_0}}=\frac{t}{\frac{{\lambda }_0}{{\mu }_1}}+\frac{2t}{\frac{{\lambda }_0}{{\mu }_2}}$
$\Rightarrow 3{\mu }_0={\mu }_1+2{\mu }_2$
$\Rightarrow {\mu }_1=3{\mu }_0-2{\mu }_2=3\times 1.4-2\times 1.2$
$\Rightarrow {\mu }_1=1.8$
Hence, refractive index of $B$ is ${\mu }_1=1.8$