Question

A plano-convex lens of refractive index μ₁ and focal length f₁ is kept in contact with another plano-concave lens of refractive index μ₂ and focal length f₂. If the radius of curvature of their spherical faces is R each and f₁ = 2f₂, then μ₁ and μ₂ are related as
1. 3μ₂ – 2μ₁ = 1
2. 2μ₂ – μ₁ = 1
3. 2μ₁ – μ₂ = 1
4. μ₁ + μ₂ = 3

Answer 1

Correct Answer - Option 3 : 2μ₁ – μ₂ = 1

Concept:

The lens maker’s equation is another formula used for lenses that give us a relationship between the focal length, refractive index, and radii of curvature of the two spheres used in lenses.

The lens maker’s equation for thin lenses is as given below as;

\(\frac{1}{f} = \left( {n - 1} \right)\left( {\frac{1}{{{R_1}}}-\frac{{\;1}}{{{R_2}}}} \right)\) 

Where,

f is the focal length (half the radius of curvature)

n is the refractive index of the material used

R₁ is the radius of curvature of sphere 1

R₂ is the radius of curvature of sphere 2

Calculation:

Given,

Refractive index of lens 1 = μ₁

Focal length of the lens 1 = f₁

Refractive index of lens 2 = μ₂

Focal length of the lens 2 = f₂

Radius of curvature of lens = R

 ⇒ f₁ = 2f₂

Now,

\(\Rightarrow \frac{1}{{\left| {{f_1}} \right|}} = \frac{1}{{\left| {2{f_2}} \right|}}\)       ----(1)

Using lens Maker’s formula, we get

\(\Rightarrow \frac{1}{{{f_1}}} = \left( {{\mu _1} - 1} \right)\left( {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}}} \right)\)

\(\Rightarrow \frac{1}{{{f_2}}} = \left( {{\mu _2} - 1} \right)\left( {\frac{1}{{{R_1}'}} - \frac{1}{{{R_2}'}}} \right)\)

Substituting these values in equation (1), we get,

\(\Rightarrow \frac{1}{{\left| {{f_1}} \right|}} = \frac{1}{{\left| {2{f_2}} \right|}}\)

\(\Rightarrow \left| {\left( {{\mu _1} - 1} \right)\left( {\frac{1}{\infty } - \frac{1}{{ - R}}} \right)} \right| = \left| {\frac{1}{2}\left( {{\mu _2} - 1} \right)\left( {\frac{1}{{ - R}} - \frac{1}{\infty }} \right)} \right|\)

\(\because\left[ {\frac{1}{\infty } = \frac{1}{{\frac{1}{0}}} = 0} \right]\)

\(\Rightarrow \left| {\left( {{\mu _1} - 1} \right)\left( {0 - \frac{1}{{ - R}}} \right)} \right| = \left| {\frac{1}{2}\left( {{\mu _2} - 1} \right)\left( {\frac{1}{{ - R}} - 0} \right)} \right|\)

\(\Rightarrow \left| {\left( {{\mu _1} - 1} \right)\left( {\frac{1}{R}} \right)} \right| = \left| {\frac{1}{2}\left( {{\mu _2} - 1} \right)\left( {\frac{1}{{ - R}}} \right)} \right|\)

\(\Rightarrow \left| {\left( {{\mu _1} - 1} \right)} \right| = \left| { - \frac{1}{2}\left( {{\mu _2} - 1} \right)} \right|\)

\(\Rightarrow \left( {{\mu _1} - 1} \right) = \frac{1}{2}\left( {{\mu _2} - 1} \right)\)

⇒ 2(μ₁ – 1) = (μ₂ – 1)

⇒ 2μ₁ – 2 = μ₂ – 1

⇒ 2μ₁ – μ₂ = -1 + 2

∴ 2μ₁ – μ₂ = 1

Therefore, the refractive indices μ₁ and μ₂ are related as 2μ₁ – μ₂ = 1.