Question

The radious of curvature of each surface of a convex lens of refractive index 1.5 is 0.4 m.Calculate the power?​

Answer 1

Explanation:

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shunutiii

Secondary School Physics 13+7 pts

The radius of curvature for a convex lens is 40 cm each surface its refractive index is 1.5 the focal length will be

Report by Homie11 18.10.2018

Answers

SerenaBochenekAmbitious

Let the radius of curvature of the convex lens are denoted as R_{1}\ and\ R_{2}\ respectively

For a convex lens R_{1} =+40\ cm [As it is towards right]

R_{2} =\ -40\ cm [As it is towards left]

The refractive index of the lens [\mu]=1.5

Let f is the focal length of the lens.

From lens maker's formula we know that -

\frac{1}{f} =[\mu -1][\frac{1}{R_{1}}- \frac{1}{ R_{2}}]

\frac{1}{f} =[1.5-1][\frac{1}{40}- \frac{1}{(-40)} ]

\frac{1}{f} =0.5*\frac{2}{40}

\frac{1}{f} =\frac{1}{40}

f=40\ cm [ans]

Hence, the focal length of the lens is 40 cm.

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gadakhsanket

Gadakhsanket★ Brainly Teacher ★

Hello friend,

For any convex lens there are two surfaces i.e. left and right

Given -

For right surface R1 = 40cm = 0.4m

For left surface R2 = -40cm = -0.4m

µ = 1.5

f = ?

Solution -

By lens makers formula,

1/f = (µ-1) (1/R1 - 1/R2)

1/f = (1.5-1)(1/0.4 - 1/-0.4)

1/f = 0.5 × (2/0.4)

f = 0.4m = 40 cm

Focal length of lens is 40 cm.

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