Double convex lens are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20 cm ?

Question

Double convex lens are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20 cm ? (A) 40 cm (B) 22 cm (C) 20 cm (D) 18 cm

Tardigrade Admin · Accepted Answer

(1/f)=(μ-1)[(1/R1)-(1/R2)] (1/20)=(1.55-1)[(1/R)-(1/(-R))] (1/20)=(0.55 × 2/R) ⇒ R=22 cm

Q. Double convex lens are to be manufactured from a glass of refractive index $1.55,$ with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be $20 c m ?$

40 cm

22 cm

20 cm

18 cm

$\frac{1}{f} = (μ - 1) [\frac{1}{R _{1}} - \frac{1}{R _{2}}]$
$\frac{1}{20} = (1.55 - 1) [\frac{1}{R} - \frac{1}{( - R )}]$
$\frac{1}{20} = \frac{0.55 \times 2}{R} \Rightarrow R = 22 c m$

Q. Double convex lens are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20cm?