MP Board · Class 12 · Physics · Ray Optics and Optical InstrumentsDerive the Lens Maker's formula for a thin convex lens, stating clearly the sign conventions and assumptions used. Also, solve a numerical problem: A convex lens of refractive index 1.5 has radii of curvature 20 cm each. Calculate its focal length in air and when immersed in water (refractive index of water = 4/3).

Derivation of Lens Maker's Formula:

Assumptions:

1. The lens is very thin so that the distances measured from its poles can be taken as equal to the distances from its optical center.
2. The aperture of the lens is small.
3. The object is a point object placed on the principal axis and the rays are paraxial.

Sign Conventions:

1. All distances are measured from the optical center of the lens.
2. Distances measured in the direction of incident light are taken as positive, and those in the opposite direction are taken as negative.

Derivation:\nLet us consider a thin convex lens made of glass of refractive index $n_2$ placed in a medium of refractive index $n_1$. Let $R_1$ and $R_2$ be the radii of curvature of the first and second refracting surfaces, respectively. \nFor the first refraction at the convex surface (refractive index $n_1$ to $n_2$): $$\frac{n_2}{v_1} - \frac{n_1}{u} = \frac{n_2 - n_1}{R_1}$__ (1) \nFor the second refraction at the concave surface (refractive index $n_2$ back to $n_1$), the image formed by the first surface acts as a virtual object: $$\frac{n_1}{v} - \frac{n_2}{v_1} = \frac{n_1 - n_2}{R_2} = -\frac{n_2 - n_1}{R_2}$__ (2) \nAdding equations (1) and (2): $$\frac{n_1}{v} - \frac{n_1}{u} = (n_2 - n_1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$$ \nDividing both sides by $n_1$ and setting $n_{21} = \frac{n_2}{n_1}$: $$\frac{1}{v} - \frac{1}{u} = \left( \frac{n_2}{n_1} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$$ \nBy definition, when $u = \infty$, $v = f$: $$\frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$$\nThis is the required Lens Maker's Formula.

Numerical Problem:

Given data:

• Refractive index of lens material ($n_g$) = 1.5
• Refractive index of air ($n_a$) = 1.0
• Refractive index of water ($n_w$) = 4/3 = 1.33
• Radius of curvature of first surface ($R_1$) = +20 cm
• Radius of curvature of second surface ($R_2$) = -20 cm

Part 1: Focal length in air ($f_a$) $$\frac{1}{f_a} = \left( \frac{n_g}{n_a} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$$ $$\frac{1}{f_a} = (1.5 - 1) \left( \frac{1}{20} - \frac{1}{-20} \right)$$ $$\frac{1}{f_a} = 0.5 \left( \frac{1}{20} + \frac{1}{20} \right) = 0.5 \left( \frac{2}{20} \right) = 0.5 \times \frac{1}{10} = \frac{1}{20}$$ $$f_a = +20\text{ cm}$$

Part 2: Focal length in water ($f_w$) $$\frac{1}{f_w} = \left( \frac{n_g}{n_w} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$$ $$\frac{1}{f_w} = \left( \frac{1.5}{4/3} - 1 \right) \left( \frac{1}{20} - \frac{1}{-20} \right)$$ $$\frac{1}{f_w} = \left( \frac{4.5}{4} - 1 \right) \left( \frac{2}{20} \right) = \left( \frac{1.125 - 1}{1} \right) \times \frac{1}{10} = 0.125 \times \frac{1}{10} = \frac{1}{80}$$ $$f_w = +80\text{ cm}$$

Answer:\nThe focal length of the lens in air is 20 cm and when immersed in water, it is 80 cm.