An equiconvex lens has a focal length \( \frac{2}{3} \) times the radius of curvature of either surface. The refractive index of the material of the lens is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 1.75
Use the lens maker's formula with the sign convention in which a surface curved towards the incoming light has a positive radius. For an equiconvex lens, \( R_1 = +R \) and \( R_2 = -R \).
\( \displaystyle \frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = (\mu - 1)\,\frac{2}{R} \)
Given \( f = 2R/3 \), we have \( 1/f = 3/(2R) \).
\( \displaystyle \frac{3}{2R} = \frac{2(\mu - 1)}{R} \;\Rightarrow\; \mu - 1 = \frac{3}{4} \;\Rightarrow\; \mu = 1.75 \)
Note: A similar question was also asked in CUET (UG) 2026 on 30-05-2026 (Slot 2).