The radii of curvature of the faces of a double convex lens of focal length 12 cm made up of glass (μ = 1.5) are 10 cm and 'p' cm respectively. The value of p is :

15 cm

The correct answer is Option (2) → 15 cm

$\frac{1}{f}=(μ-1)(\frac{1}{R_1}-\frac{1}{R_2})$ [lens maker formula]

where,

$f$ = focal length = 12 cm [given]

Refractive index of glass, $μ=1.5$ [given]

$R_1$ = Radius of curvature of 1st surface = 10 cm

$R_2$ = Radius of curvature of 1st surface = -p cm

[since the second surface is concave]

$∴\frac{1}{12}=0(1.5-1)(\frac{1}{10}-\frac{1}{-p})$

$\frac{1}{6}=\frac{1}{10}+\frac{1}{p})$

$⇒p=15cm$