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

Question

Double convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of same radius of curvature. What is the radius of curvature required if the focal length is to be 20 cm? (A) 11 cm (B) 22 cm (C) 7 cm (D) 6 cm

Tardigrade Admin · Accepted Answer

Lens maker s formula, (1/f) = (μ-1)((1/R1)-(1/R2)) Here, f = 20 cm, μ= 1.55, R1 =R and R2 = -R ∴ (1/20)=(1.55-1)((1/R )-(1/(-R))) or (1/20)=0.55×(2/R) ⇒ R = 1.1 × 20 = 22 cm

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

$11 c m$

$22 c m$

$7 c m$

$6 c m$

Lens maker s formula, $\frac{1}{f} = (μ - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$
Here, $f = 20 c m, μ = 1.55$ ,
$R_{1} = R$ and $R_{2} = - R$
$∴ \frac{1}{20} = (1.55 - 1) (\frac{1}{R} - \frac{1}{( - R )})$ or
$\frac{1}{20} = 0.55 \times \frac{2}{R}$
$\Rightarrow R = 1.1 \times 20$
$= 22 c m$

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

11cm

22cm

7cm

6cm