Question

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 1

✦AnSwer:

Focal length of the lens is 100 cm.

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✩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.

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