The power of a biconvex lens is 10 dioptre and the radius of curvature of each surface is 10 cm. Then the refractive index of the material of the lens is,

Question

The power of a biconvex lens is 10 dioptre and the radius of curvature of each surface is 10 cm. Then the refractive index of the material of the lens is, (A) (4/3) (B) (9/8) (C) (5/3) (D) (3/2)

Tardigrade Admin · Accepted Answer

P=(100/f) ⇒ f=(100/P) =(100/10)=10 cm f=(R/2(μ-1)) (for equiconvex lens) 10=(10/2(μ-1)) (μ-1)=(1/2) ⇒ μ=(1/2)+1 =(3/2)

Q. The power of a biconvex lens is $10$ dioptre and the radius of curvature of each surface is $10 c m$ . Then the refractive index of the material of the lens is,

$\frac{4}{3}$

$\frac{9}{8}$

$\frac{5}{3}$

$\frac{3}{2}$

$P = \frac{100}{f}$
$\Rightarrow f = \frac{100}{P}$
$= \frac{100}{10} = 10 c m$
$f = \frac{R}{2 ( μ - 1 )}$ (for equiconvex lens)
$10 = \frac{10}{2 ( μ - 1 )}$
$(μ - 1) = \frac{1}{2}$
$\Rightarrow μ = \frac{1}{2} + 1$
$= \frac{3}{2}$

Q. The power of a biconvex lens is 10 dioptre and the radius of curvature of each surface is 10cm. Then the refractive index of the material of the lens is,

34​

89​

35​

23​