The radius of curvature of spherical surfaces of a convex lens are 15

1.	The radius of curvature of spherical surfaces of a convex lens are 15 cm. and 10 cm.respectively. If Refractive Index of lens is 1.5. then find its focal length.
Answer» ✦AnSwer: Focal length of the lens is 100 cm. ━━━━━━━━━━━━━━━━ ✩GiVen: REFRACTIVE Index = 1.5 R1 = 15 cm R2 = 10 cm ✩To Find: We need to find the focal length of CONVEX lens. ✩SoluTion: According to the formula, $\sf \longmapsto \: \: \dfrac{1}{f} = \bigg(\dfrac{n_{2}}{n_{1}} - 1 \bigg) \bigg(\dfrac{1}{r_{1}} - \dfrac{1}{r_{2}} \bigg)$ $\sf \longmapsto \: \: \dfrac{1}{f} = (1.5 - 1) \bigg( \dfrac{1}{15} - \dfrac{1}{10} \bigg)$ $\sf \longmapsto \: \: \dfrac{1}{f} = (0.5) \bigg( \dfrac{15 - 10}{150} \bigg)$ $\sf \longmapsto \: \: \dfrac{1}{f} = (0.5) \bigg( \dfrac{5}{150} \bigg)$ $\sf \longmapsto \: \: \dfrac{1}{f} = \dfrac{2.5}{150}$ $\sf \longmapsto \: \: { \underline{ \boxed{ \bf{ \purple{{ \bold{f \: = 100 \: cm}}}}}}} \: \bigstar$ THEREFORE, focal length is 100 cm. ━━━━━━━━━━━━━━━━

The radius of curvature of spherical surfaces of a convex lens are 15 cm. and 10 cm.respectively. If Refractive Index of lens is 1.5. then find its focal length.

Answer»

Focal length of the lens is 100 cm.

According to the formula,

$\sf \longmapsto \: \: \dfrac{1}{f} = \bigg(\dfrac{n_{2}}{n_{1}} - 1 \bigg) \bigg(\dfrac{1}{r_{1}} - \dfrac{1}{r_{2}} \bigg)$

$\sf \longmapsto \: \: \dfrac{1}{f} = (1.5 - 1) \bigg( \dfrac{1}{15} - \dfrac{1}{10} \bigg)$

$\sf \longmapsto \: \: \dfrac{1}{f} = (0.5) \bigg( \dfrac{15 - 10}{150} \bigg)$

$\sf \longmapsto \: \: \dfrac{1}{f} = (0.5) \bigg( \dfrac{5}{150} \bigg)$

$\sf \longmapsto \: \: \dfrac{1}{f} = \dfrac{2.5}{150}$

$\sf \longmapsto \: \: { \underline{ \boxed{ \bf{ \purple{{ \bold{f \: = 100 \: cm}}}}}}} \: \bigstar$