A double convex lens has a focal length of $20 ext{ cm}$ in air. The refractive index of the lens material is $1.5$. If this lens is completely immersed in water of refractive index $1.33$, calculate its new focal length in water. Also, state whether the nature of the lens changes.

Q: A double convex lens has a focal length of $20\text{ cm}$ in air. The refractive index of the lens material is $1.5$. If this lens is completely immersed in water of refractive index $1.33$, calculate its new focal length in water. Also, state whether the nature of the lens changes.

Given Data: Focal length of the lens in air, $fa = 20\text{ cm}$ Refractive index of the lens material with respect to air, $ng = 1.5$ Refractive index of water with respect to air, $nw = 1.33 = \frac{4}{3}$ Refractive index of air, $na = 1.0$ Formula Used:\nAccording to the Lens Maker's Formula: $$\frac{1}{f} = \left( \frac{n2}{n1} - 1 \right) \left( \frac{1}{R1} - \frac{1}{R2} \right)$$ Case 1: When the lens is in air $$\frac{1}{fa} = \left( \frac{ng}{na} - 1 \right) \left( \frac{1}{R1} - \frac{1}{R2} \right)$$ $$\frac{1}{20} = (1.5 - 1) \left( \frac{1}{R1} - \frac{1}{R2} \right)$$ $$\frac{1}{20} = 0.5 \left( \frac{1}{R1} - \frac{1}{R2} \right) \quad \text{--- (Equation 1)}$$ Case 2: When the lens is immersed in water $$\frac{1}{fw} = \left( \frac{ng}{nw} - 1 \right) \left( \frac{1}{R1} - \frac{1}{R2} \right)$$ $$\frac{1}{fw} = \left( \frac{1.5}{4/3} - 1 \right) \left( \frac{1}{R1} - \frac{1}{R2} \right)$$ $$\frac{1}{fw} = \left( \frac{4.5}{4} - 1 \right) \left( \frac{1}{R1} - \frac{1}{R2} \right)$$ $$\frac{1}{fw} = \left( \frac{4.5 - 4}{4} \right) \left( \frac{1}{R1} - \frac{1}{R2} \right)$$ $$\frac{1}{fw} = \left( \frac{0.5}{4} \right) \left( \frac{1}{R1} - \frac{1}{R2} \right) \quad \text{--- (Equation 2)}$$ Calculation:\nDividing Equation 1 by Equation 2: $$\frac{1/20}{1/fw} = \frac{0.5 \left( \frac{1}{R1} - \frac{1}{R2} \right)}{\left( \frac{0.5}{4} \right) \left( \frac{1}{R1} - \frac{1}{R2} \right)}$$ $$\frac{fw}{20} = \frac{0.5}{0.5 / 4} = 4$$ $$fw = 4 \times 20 = 80\text{ cm}$$ Nature of the Lens:\nThe new focal length of the lens in water is $+80\text{ cm}$. Since the sign of the focal length remains positive, the lens continues to behave as a converging (convex) lens. Its fundamental nature does not change, although its converging power decreases because the relative refractive index between the lens and the surrounding medium has decreased.

Question

A double convex lens has a focal length of $20	ext{ cm}$ in air. The refractive index of the lens material is $1.5$. If this lens is completely immersed in water of refractive index $1.33$, calculate its new focal length in water. Also, state whether the nature of the lens changes.

Accepted Answer

Given Data: Focal length of the lens in air, $fa = 20	ext{ cm}$ Refractive index of the lens material with respect to air, $ng = 1.5$ Refractive index of water with respect to air, $nw = 1.33 = \frac{4}{3}$ Refractive index of air, $na = 1.0$ Formula Used:
According to the Lens Maker's Formula: $$\frac{1}{f} = \left( \frac{n2}{n1} - 1 ight) \left( \frac{1}{R1} - \frac{1}{R2} ight)$$ Case 1: When the lens is in air $$\frac{1}{fa} = \left( \frac{ng}{na} - 1 ight) \left( \frac{1}{R1} - \frac{1}{R2} ight)$$ $$\frac{1}{20} = (1.5 - 1) \left( \frac{1}{R1} - \frac{1}{R2} ight)$$ $$\frac{1}{20} = 0.5 \left( \frac{1}{R1} - \frac{1}{R2} ight) \quad 	ext{--- (Equation 1)}$$ Case 2: When the lens is immersed in water $$\frac{1}{fw} = \left( \frac{ng}{nw} - 1 ight) \left( \frac{1}{R1} - \frac{1}{R2} ight)$$ $$\frac{1}{fw} = \left( \frac{1.5}{4/3} - 1 ight) \left( \frac{1}{R1} - \frac{1}{R2} ight)$$ $$\frac{1}{fw} = \left( \frac{4.5}{4} - 1 ight) \left( \frac{1}{R1} - \frac{1}{R2} ight)$$ $$\frac{1}{fw} = \left( \frac{4.5 - 4}{4} ight) \left( \frac{1}{R1} - \frac{1}{R2} ight)$$ $$\frac{1}{fw} = \left( \frac{0.5}{4} ight) \left( \frac{1}{R1} - \frac{1}{R2} ight) \quad 	ext{--- (Equation 2)}$$ Calculation:
Dividing Equation 1 by Equation 2: $$\frac{1/20}{1/fw} = \frac{0.5 \left( \frac{1}{R1} - \frac{1}{R2} ight)}{\left( \frac{0.5}{4} ight) \left( \frac{1}{R1} - \frac{1}{R2} ight)}$$ $$\frac{fw}{20} = \frac{0.5}{0.5 / 4} = 4$$ $$fw = 4 	imes 20 = 80	ext{ cm}$$ Nature of the Lens:
The new focal length of the lens in water is $+80	ext{ cm}$. Since the sign of the focal length remains positive, the lens continues to behave as a converging (convex) lens. Its fundamental nature does not change, although its converging power decreases because the relative refractive index between the lens and the surrounding medium has decreased.