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. The refractive index of the glass is $1.5$. (a) Calculate the focal length of the lens in air. (b) If this lens is immersed in water of refractive index $1.33$, calculate its new focal length in water.

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 glass ($n_g$) = $1.5$
• Refractive index of water ($n_w$) = $1.33$ or $\frac{4}{3}$
• Refractive index of air ($n_a$) = $1.0$

Part (a): Focal Length in Air ($f_a$)\nUsing the lens maker's formula:

$$\frac{1}{f_a} = \left(\frac{n_g}{n_a} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$ \nSubstituting 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 (b): Focal Length in Water ($f_w$)\nUsing the lens maker's formula for water medium:

$$\frac{1}{f_w} = \left(\frac{n_g}{n_w} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$ \nSubstitute $n_g = 1.5$ and $n_w = \frac{4}{3}$: $$\frac{n_g}{n_w} = \frac{1.5}{4/3} = \frac{3/2}{4/3} = \frac{9}{8} = 1.125$$ \nNow substitute into the equation: $$\frac{1}{f_w} = \left(\frac{9}{8} - 1\right) \left(\frac{1}{20} - \frac{1}{-30}\right)$$ $$\frac{1}{f_w} = \left(\frac{1}{8}\right) \left(\frac{5}{60}\right) = \frac{5}{480} = \frac{1}{96}$$ $$f_w = +96\text{ cm}$$

Conclusion:

• Focal length of the lens in air is $24\text{ cm}$.
• Focal length of the lens in water is $96\text{ cm}$.