A double convex thin lens made of glass (refractive index μ=1.5 ) has both radii of curvature of magnitude 20 cm . Incident light rays parallel to the axis of the lens will converge at a distance L such that

Question

A double convex thin lens made of glass (refractive index μ=1.5 ) has both radii of curvature of magnitude 20 cm . Incident light rays parallel to the axis of the lens will converge at a distance L such that (A) L = 20 cm (B) L = 10 cm (C) L = 40 cm (D) L = 20 / 3 cm

Tardigrade Admin · Accepted Answer

(1/f)=(μ-1)((1/R1)-(1/R2)) For double convex lens, μ=1.5, R1=+20 cm , R2=-20 (1/f)=(1.5-1)((1/(+20))-(1/(-20))) (1/f)=0.5[(2/20)] ⇒ f=+20 cm

Q. A double convex thin lens made of glass (refractive index $μ = 1.5$ ) has both radii of curvature of magnitude $20 c m .$ Incident light rays parallel to the axis of the lens will converge at a distance $L$ such that

L = 20 cm

L = 10 cm

L = 40 cm

L = 20 / 3 cm

$\frac{1}{f} = (μ - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$
For double convex lens,
$μ = 1.5, R_{1} = + 20 c m, R_{2} = - 20$
$\frac{1}{f} = (1.5 - 1) (\frac{1}{( + 20 )} - \frac{1}{( - 20 )})$
$\frac{1}{f} = 0.5 [\frac{2}{20}] \Rightarrow f = + 20 c m$

Q. A double convex thin lens made of glass (refractive index μ=1.5 ) has both radii of curvature of magnitude 20cm. Incident light rays parallel to the axis of the lens will converge at a distance L such that