JEE Advanced · 2026 · 17 May 2026 · Paper 1 · Geometrical Optics

Using the lens maker's formula: $\frac{1}{\mathrm{f}}=\left(\frac{1.5}{{\mathrm{n}}_L}-1\right)\left(\frac{1}{0.2}-\frac{1}{-0.2}\right)=\left(\frac{1.5}{{\mathrm{n}}_L}-1\right)\cdot 10$

According to NCERT convention, power $\mathrm{P}=\frac{1}{\mathrm{f}}$ (in meters).

Therefore: $\mathrm{P}=\frac{15}{{\mathrm{n}}_L}-10$

This is a rectangular hyperbola. When $n_L=1.5$, $P=0$ (lens becomes invisible). For $n_L<1.5$, power is positive (converging), and for $n_L>1.5$, power is negative (diverging).

Alternatively, if $\mathrm{P}=\frac{{\mathrm{n}}_L}{\mathrm{f}}$, then $\mathrm{P}=(1.5-{\mathrm{n}}_L)\cdot 10$, which is a straight line.

Both interpretations exist in different contexts, making options (A) or (B) potentially correct.