A double convex lens has two surfaces of equal radii 'R' and refractive index $n = 2.5$ . Find the focal length $^{'} f^{'}$

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Solution

Concept: Lens maker's formula
$\frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
Solutio: From lens maker's formula and according to question
$\frac{1}{f} = (2.5 - 1) (\frac{1}{R_{1}} + \frac{1}{R_{2}})$
$\frac{1}{f} = \frac{3}{2} / \frac{2}{R}$
$f = \frac{R}{3}$

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A double convex lens has two surfaces of equal radii 'R' and refractive index n=2.5. Find the focal length ′f′

Concept: Lens maker's formula1f=(μ−1)(1R1−1R2)Solutio: From lens maker's formula and according to question1f=(2.5−1)(1R1+1R2)1f=32/2Rf=R3

A double convex lens has two surfaces of equal radii 'R' and refractive index $n = 2.5$ . Find the focal length $^{'} f^{'}$

Concept: Lens maker's formula
$\frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
Solutio: From lens maker's formula and according to question
$\frac{1}{f} = (2.5 - 1) (\frac{1}{R_{1}} + \frac{1}{R_{2}})$
$\frac{1}{f} = \frac{3}{2} / \frac{2}{R}$
$f = \frac{R}{3}$