JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Geometrical Optics

The ray refracts at the first face through angle $r_1$, then meets the coated back face at angle $r_2$, with $r_1+r_2=A=75^{\circ }$. Total internal reflection at the back face needs $r_2$ to exceed the critical angle of the glass–film boundary, where $\sin {\theta }_c=\frac{n_{film}}{n_{glass}}=\frac{1.5}{5/3}=0.9$, so ${\theta }_c\approx 64.2^{\circ }$. At the limit $r_2={\theta }_c$, $r_1=75^{\circ }-64.2^{\circ }=10.8^{\circ }$, and the first face gives $\sin i=n_{glass}\sin r_1=\frac{5}{3}\sin 10.8^{\circ }\approx \frac{5}{3}(0.188)=0.313$, i.e. $i\approx 18.3^{\circ }$. A larger $r_2$ (stronger TIR) corresponds to a smaller $r_1$ and hence a smaller $i$, so TIR occurs for incidence below about $18.3^{\circ }$. Checking the given data, $i=15^{\circ }$ gives $r_1\approx 8.6^{\circ }$ and $r_2\approx 66.4^{\circ }>{\theta }_c$ (TIR occurs), while $i=25^{\circ }$ gives $r_1\approx 15.0^{\circ }$ and $r_2\approx 60.0^{\circ }<{\theta }_c$ (no TIR); the limiting incident angle lies between $15^{\circ }$ and $20^{\circ }$.