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Lens makers formula

Lens maker’s formula

Lens makers formula :- Lens maker’s formula is the relation between the focal length of a lens, the refractive index of its material and the radii of curvature of its two surfaces. It is used by lens manufacturers to make the lenses of particular power from the glass of a given refractive index.

Let us suppose that a double convex lens made of glass of refractive index n_Lis placed in a rarer medium of refractive index n_M. Below figure shows the geometry of image formation by the lens.

The image formation can be seen in terms of two steps :-
(i) Formation of image I₁ of the object O by the first refracting surface XP₁Y

If the lens material were continuous and there were no boundary/second surface XP₂Y of the lens, then the refraction would take place only at surface XP₁Y and the refracted ray would go straight meeting the ray going along principle axis at point I₁. Therefore I₁ would have been a real image of O formed after refraction at XP₁Y as shown in figure below :-

Now using the formula of refraction at spherical surface, i.e.,

$\displaystyle \frac{{{n}_{\text{material of refracted ray}}}}{v}-\frac{{{n}_{\text{material of incident ray}}}}{u}=\frac{{{n}_{\text{material of refracted ray}}}-{{n}_{\text{material of incident ray}}}}{R}$

Here n_{material of refracted ray} = n_Land n_{material of incident ray} = n_M

Object distance P₁O = u and image distance P₁I₁ = v₁

So we get $\displaystyle \frac{{{n}_{L}}}{{{v}_{1}}}-\frac{{{n}_{M}}}{u}=\frac{{{n}_{L}}-{{n}_{\text{M}}}}{{{R}_{1}}}$ …..(1)

(ii) Formation of image I of the virtual object I₁ by the second refracting surface XP₂Y

Actually the lens material is not continuous. Therefore the refracted ray again suffers refraction at surface XP₂Y of the lens, meeting the principle axis actually at I. Therefore I is the final real image of O, formed after refraction through the convex lens as shown in figure below :-

For refraction at the second surface XP₂Y, we can consider I₁ as a virtual object, whose real image is formed at I.

Again using

Now n_{material of refracted ray} = n_Mand n_{material of incident ray} = n_L

Object distance P₂I₁ ≅ P₁I₁ = v₁ and image distance P₂I = v

So we get $\displaystyle \frac{{{n}_{M}}}{v}-\frac{{{n}_{L}}}{{{v}_{1}}}=\frac{{{n}_{M}}-{{n}_{\text{L}}}}{{{R}_{2}}}$ …..(2)

Adding equation (1) and (2), we get

$\displaystyle \left[ \frac{{{n}_{L}}}{{{v}_{1}}}-\frac{{{n}_{M}}}{u} \right]+\left[ \frac{{{n}_{M}}}{v}-\frac{{{n}_{L}}}{{{v}_{1}}} \right]=\left[ \frac{{{n}_{L}}-{{n}_{\text{M}}}}{{{R}_{1}}} \right]+\left[ \frac{{{n}_{M}}-{{n}_{\text{L}}}}{{{R}_{2}}} \right]$

$\displaystyle \Rightarrow \frac{{{n}_{M}}}{v}-\frac{{{n}_{M}}}{u}=\left[ \frac{{{n}_{L}}-{{n}_{\text{M}}}}{{{R}_{1}}} \right]-\left[ \frac{{{n}_{L}}-{{n}_{\text{M}}}}{{{R}_{2}}} \right]$

$\displaystyle {{n}_{M}}\left( \frac{1}{v}-\frac{1}{u} \right)=\left( {{n}_{L}}-{{n}_{\text{M}}} \right)\left[ \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right]$

$\displaystyle \Rightarrow \left( \frac{1}{v}-\frac{1}{u} \right)=\left( \frac{{{n}_{L}}}{{{n}_{M}}}-1 \right)\left[ \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right]$

Now put n_L/n_M = n = refractive index of material of the lens w.r.t. surrounding medium.

So $\displaystyle \Rightarrow \left( \frac{1}{v}-\frac{1}{u} \right)=\left( n-1 \right)\left[ \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right]$ …..(3)

When object on the left of lens is at infinity, image is formed at the principle focus of the lens.

∴ when u = ∞, v = f = focal length of the lens.

From equation (3),

$\displaystyle \left( \frac{1}{f}-\frac{1}{\infty } \right)=\left( n-1 \right)\left[ \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right]$

$\displaystyle \frac{1}{f}=\left( n-1 \right)\left[ \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right]$

This is the Lens Makers Formula.

Note :-

(1). For a convex lens R₁ is positive and R₂ is negative.

So, $\displaystyle \frac{1}{f}=\left( n-1 \right)\left[ \frac{1}{+{{R}_{1}}}-\frac{1}{-{{R}_{2}}} \right]$

$\displaystyle \Rightarrow \frac{1}{f}=\left( n-1 \right)\left[ \frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}} \right]$

⇒ focal length of convex lens is positive.

(2). For a concave lens R₁ is negative and R₂ is positive.

$\displaystyle \frac{1}{f}=\left( n-1 \right)\left[ \frac{1}{-{{R}_{1}}}-\frac{1}{+{{R}_{2}}} \right]$

$\displaystyle \Rightarrow \frac{1}{f}=-\left( n-1 \right)\left[ \frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}} \right]$

⇒ focal length of concave lens is negative

3. Lens immersed in a liquid

Suppose that focal length of lens in air is f_a and in liquid is f_l, then

$\displaystyle \frac{1}{{{f}_{a}}}=\left( {{n}_{ga}}-1 \right)\left[ \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right]$

$\displaystyle \frac{1}{{{f}_{l}}}=\left( {{n}_{gl}}-1 \right)\left[ \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right]$

Now $\displaystyle \frac{{{f}_{l}}}{{{f}_{a}}}=\frac{\left( {{n}_{ga}}-1 \right)}{\left( {{n}_{gl}}-1 \right)}$

Here $\displaystyle {{n}_{gl}}=\frac{{{n}_{g}}}{{{n}_{l}}}$

(A) If a lens is immersed in a liquid whose refractive index is less than refractive index of substance of lens, then

$\displaystyle {{n}_{gl}}=\frac{{{n}_{g}}}{{{n}_{l}}}>1$

& $\displaystyle {{n}_{ga}}>{{n}_{gl}}$

Then f_l > f_a

So the focal length increases but its nature does not change.

(B) If a lens is immersed in a liquid whose refractive index is equal to the refractive index of substance of lens, then

$\displaystyle {{n}_{gl}}=\frac{{{n}_{g}}}{{{n}_{l}}}=1$

⇒ f_l= ∞ and P_l = 0

It means focal length of lens will become infinite and it will act like a flat transparent plate.

(C) If a lens is immersed in a liquid whose refractive index is greater than refractive index of substance of lens, then

$\displaystyle {{n}_{gl}}=\frac{{{n}_{g}}}{{{n}_{l}}}<1$

then the value of f_l will be negative

It means the convex lens in the air will behave like a concave lens in liquid. The nature of the lens will change that is why, the air bubble (convex) in water behaves like a concave lens.

(D) The focal length of lens (n =1.5) is f then it’s focal length will become 4f, when immersed in water(refractive index of water is 4/3).

(E) From $\displaystyle \frac{1}{f}=\left( n-1 \right)\left[ \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right]$ , we get

$\displaystyle \Rightarrow f\propto \frac{1}{n-1}$

⇒ focal length of lens is the highest for red color and lowest for violet color.

Next Topic :- Sign Convention For Lens

Lens makers formula | Lens maker’s formula

Lens makers formula

Lens maker’s formula

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