Convex lens made up of glass μ_g = 1.5 and radius of curvature R is dipped into the water. Its focal length will be (Refractive index of water = 43)

Question

MARKS Content Team · Accepted Answer

Correct answer: (A) 4R. Here, μ_g=1.5=3/2, μ_w ᷿=4/3 R_1=R,⇒R_2=-R From, lens maker's formula 1/f=μ-11/R_1-1/R_2where μ=(μ_L)/(μ_M) ∴1/(f_aₐ)=(μ_g)/(μ_aₐ)-11/R-1/(-R) ⇒1/f_a=1/R Also 1/(f_w ᷿)=(μ_g)/(μ_wₐ)-11/R-1/(-R) Or 1/f_w=1/4R Or f_w ᷿=4R

Convex lens made up of glass $(μ_{g} = 1.5)$ and radius of curvature $R$ is dipped into the water. Its focal length will be (Refractive index of water = $\frac{4}{3}$ )

Solution

More Ray Optics Questions

Convex lens made up of glass μg = 1.5 and radius of curvature R is dipped into the water. Its focal length will be (Refractive index of water = 43)

Solution

More Ray Optics Questions

Convex lens made up of glass $(μ_{g} = 1.5)$ and radius of curvature $R$ is dipped into the water. Its focal length will be (Refractive index of water = $\frac{4}{3}$ )