MP Board · Class 12 · Physics · Ray Optics and Optical InstrumentsA double convex lens has radii of curvature $20\text{ cm}$ and $30\text{ cm}$ respectively. If the refractive index of the material of the lens is $1.5$, calculate its focal length. What will be the new focal length if the lens is immersed in water of refractive index $1.33$?

Given Data:

• Radius of curvature of first surface ($R_1$) = $+20\text{ cm}$
• Radius of curvature of second surface ($R_2$) = $-30\text{ cm}$
• Refractive index of lens ($n_g$) = $1.5$
• Refractive index of water ($n_w$) = $1.33$ or $\frac{4}{3}$

Part 1: Focal length in air ($f_a$)\nUsing the Lens Maker's Formula:

$$\frac{1}{f_a} = (n_g - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$\nSubstitute the given values: $$\frac{1}{f_a} = (1.5 - 1)\left(\frac{1}{20} - \frac{1}{-30}\right)$$ $$\frac{1}{f_a} = 0.5 \left(\frac{1}{20} + \frac{1}{30}\right)$$ $$\frac{1}{f_a} = 0.5 \left(\frac{3 + 2}{60}\right) = 0.5 \left(\frac{5}{60}\right) = \frac{0.5}{12} = \frac{1}{24}$$ $$f_a = +24\text{ cm}$$

Part 2: Focal length in water ($f_w$)\nWhen the lens is immersed in water, the relative refractive index is $n_{gw} = \frac{n_g}{n_w} = \frac{1.5}{4/3} = \frac{4.5}{4} = 1.125$.\nUsing the Lens Maker's Formula in water:

$$\frac{1}{f_w} = (n_{gw} - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$ $$\frac{1}{f_w} = (1.125 - 1)\left(\frac{1}{20} - \frac{1}{-30}\right)$$ $$\frac{1}{f_w} = 0.125 \left(\frac{5}{60}\right) = \frac{1}{8} \times \frac{1}{12} = \frac{1}{96}$$ $$f_w = +96\text{ cm}$$

Conclusion:\nThe focal length of the lens in air is $24\text{ cm}$ and when immersed in water, it increases to $96\text{ cm}$.