Question

A double convex lens of equal radii of curvature 'R' and refractive index of material is 1.5 .What is focal length?​

Answer 1

Answer:

$\frac{1}{f}$ = (n-1)($\frac{1}{R1}$ +$\frac{1}{R2}$)

n=1.5

R1=R2=R

$\frac{1}{f}$ =(1.5-1)(1/R+1/R)

1/f=(0.5)(1+1/R)

1/f=1/2(2/R)

1/f=1/R

f=R

Answer 2

Given: the refractive index of the material, μ = 1.5

           the radii of curvature of the lens are  R₁ and R₂ = R

To Find: the focal length of the double convex lens, f.

Solution:

To calculate f, the formula used:

• 1 / f = (μ - 1) x [ 1/R₁ - 1/R₂]                          ⇒ Lens maker's formula

Applying the above formula

1 / f =  (μ - 1) x [ 1/R₁ - (-1/R₂)]  

     =  (1.5 - 1) x [ 1/R₁ - (-1/R₂)]

Negative sign of R₂ is as per the sign convention rule.

As R₁ and R₂ are equal i,e. R₁ = R₂ = R

∴ 1 / f = (1.5 - 1) x [ 1/R - (-1/R)]

         = 0.5 x [ 1/R + 1/R]

         = 0.5 x [ 2 / R]

         = 0.5 x 2 / R

         = 1.0 / R

         = 10 / 10xR

         = 1 / R

1 / f    = 1 / R

or, f   =  R

Hence, the focal length of the convex lens is equal to its radius of curvature.