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?

Toppr Admin · Accepted Answer

Lens maker-apos-s formula-1f-x3BC-x2212-1-1R1-x2212-1R2-Here- f - 20 cm- -x3BC- - 1-55- R1 - R- R2 - -R120-1-55-x2212-1-1R-x2212-1-x2212-R-or120-0-55-xD7-2R-x21D2-R-1-1-xD7-20-22cm

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?

Lens maker's formula,1f=(μ−1)(1R1−1R2)Here, f = 20 cm, μ = 1.55, R1 = R, R2 = -R120=(1.55−1)(1R−1(−R))or120=0.55×2R⇒R=1.1×20=22cm

Lens maker's formula,
$\frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
Here, f = 20 cm, $μ$ = 1.55, $R_{1}$ = R, $R_{2}$ = -R
$\frac{1}{20} = (1.55 - 1) (\frac{1}{R} - \frac{1}{(- R)}) o r \frac{1}{20} = 0.55 \times \frac{2}{R}$
$\Rightarrow R = 1.1 \times 20 = 22 c m$