Question

Consider slabs of three media A,B and C arranged as shown in figure R.I of A is 1.5 and that of C is 1.4. If the number of waves in A is equal to the number of waves in the combination of B and C then refractive index of B is:

• A. 1.4
• B. 1.5
• C. 1.6
• D. 1.7

Answer 1

The correct option is D 1.7

If ${\lambda }_0$ be the wavelength outside the slab, then we can say that

${\mu }_{A,B,C}=\frac{{\lambda }_0}{{\lambda }_{A,B,C}}$

$\therefore {\lambda }_A=\frac{{\lambda }_0}{{\mu }_A}$ ${\lambda }_B=\frac{{\lambda }_0}{{\mu }_B}$ ${\lambda }_C=\frac{{\lambda }_0}{{\mu }_C}$

Number of waves$=\frac{d}{\lambda }$

From the figure we have the distances

$d_A=3x$ $d_B=x$ $d_C=2x$

$\therefore \frac{d_A}{{\lambda }_A}=\frac{d_B}{{\lambda }_B}+\frac{d_C}{{\lambda }_C}$

$\Rightarrow \frac{3x}{{\lambda }_0/{\mu }_A}=\frac{x}{{\lambda }_0/{\mu }_B}+\frac{2x}{{\lambda }_0/{\mu }_C}$

$\Rightarrow 3{\mu }_A={\mu }_B+2{\mu }_C$

$\Rightarrow {\mu }_B=3{\mu }_A-2{\mu }_C$

$\Rightarrow {\mu }_B=3\times 1.5-2\times 1.4=1.7$

Hence the correct answer is option (D).