A convex lens of refractive index 1.5 and focal length f = 18 cm is immersed in water. The difference in focal lengths of the given lens when it is in water and in air is α × f. The value of α i $s ________.$
(refractive index of water = 4/3)

Question

A convex lens of refractive index 1.5 and focal length f = 18 cm is immersed in water. The difference in focal lengths of the given lens when it is in water and in air is α × f. The value of α i $s ________.$

(refractive index of water = 4/3)

G1 Physics · Accepted Answer

For a thin lens in a medium, 1/f ∝ (μ(relative)-1), where μ(relative) = μ(lens)/μ(medium). In air, μ(relative,air) = 1.5. Therefore, μ(relative,air)-1 = 0.5. In water, μ(relative,water) = 1.5/(4/3) = 9/8 = 1.125. Hence, μ(relative,water)-1 = 0.125 = 1/8. Therefore, f(water)/f(air) = 0.5/0.125 = 4. Thus, f(water) = 4f. The difference is f(water)-f(air) = 4f-f = 3f. The question writes the difference as αf. Therefore, α = 3. Hence the answer is 3.

A convex lens of refractive index 1.5 and focal length f = 18 cm is immersed in water. The difference in focal lengths of the given lens when it is in water and in air is α × f. The value of α i s ________. (refractive index of water = 4/3)