GATE 2023 EC Q.60: solution and answer

Official answer: 0.12 to 0.14Source: Official answer key, GATE 2023 EC (IIT Kanpur)

1. Refractive indices. For a lossless medium, $n=\sqrt{{\mu }_r{\epsilon }_r}$. Air: $n_1=1$. Glass: $n_3=\sqrt{1\times 4}=2$.

2. Condition for zero reflection (quarter-wave matching). At normal incidence from air onto glass through a single coating layer, reflection is zero only when the coating acts as a quarter-wave transformer between the two media. In impedance form this is ${\eta }_2=\sqrt{{\eta }_1{\eta }_3}$. For non-magnetic media $\eta ={\eta }_0/n$ (with ${\eta }_0\approx 377\Omega$), so $$n_2=\sqrt{n_1{n}_3}=\sqrt{1\times 2}=\sqrt{2}\approx 1.414({\epsilon }_{r2}=2)$$

3. Thickness condition. The coating thickness must be an odd multiple of a quarter of the wavelength inside the coating: $$d=(2m+1)\frac{{\lambda }_2}{4},{\lambda }_2=\frac{{\lambda }_0}{n_2},m=0,1,2,\dots$$

4. Minimum thickness (take $m=0$): $$d_{\min }=\frac{{\lambda }_0}{4n_2}=\frac{0.75\mu \text{m}}{4\sqrt{2}}=\frac{0.75}{5.657}\mu \text{m}\approx 0.1326\mu \text{m}$$

5. Check. The wave reflected at the coating–glass face travels an extra distance $2d={\lambda }_2/2$, which adds a phase of $2{\beta }_{2}d=\pi$. Both interfaces go from lower to higher index, so both reflections pick up the same sign. The two reflected waves therefore arrive out of phase. Because $n_2=\sqrt{n_1{n}_3}$, the two interface reflection coefficients are equal: $\frac{1-\sqrt{2}}{1+\sqrt{2}}$ at air–coating and $\frac{\sqrt{2}-2}{\sqrt{2}+2}=\frac{1-\sqrt{2}}{1+\sqrt{2}}$ at coating–glass. The exact transmission-line result agrees: with $d={\lambda }_2/4$, the impedance seen from the air side is ${\eta }_{in}={\eta }_2^2/{\eta }_3={\eta }_1$, so the net reflection is zero.

6. Rounded to two decimal places, $d_{\min }\approx 0.13\mu \text{m}$.