A double-convex lens of power P, with each face having same radius of curvature, is cut into two equal parts perpendicular to its principal axis. The power of one part of the lens will be: (A)2P (B) P (C) 4P (D) P/2

n₁ = refractive index of medium outside the lens (usually assumed as air with n = 1)
n₂ = refractive index of lens
R₁ = radius of side₁ of lens
R₂ = radius of side₂ of lens

Given:

for double-convex lens: R₁ = R₂ = R
for cut lens: R₁ = R

R₂ = ∞ (flat surface has a radius of ∞)

[tex]\displaystyle P_{double-convex}:P_{cut\ lens}[/tex]

[tex]\displaystyle =\left[\left(\frac{n_2}{n_1}-1\right)\left(\frac{1}{R} +\frac{1}{R} \right) \right]:\left[\left(\frac{n_2}{n_1}-1\right)\left(\frac{1}{R} +\frac{1}{\infty} \right)\right][/tex]

[tex]\displaystyle =\left(\frac{1}{R} +\frac{1}{R} \right):\left(\frac{1}{R} +\frac{1}{\infty} \right)[/tex]

[tex]\displaystyle =\frac{2}{R} :\left(\left\frac{1}{R} +0\right)[/tex]

[tex]=2:1[/tex]

Therefore, the power of the cut lens will be [tex]\frac{1}{2}[/tex] of the power of the double-convex lens.

The answer is (D) P/2

A double-convex lens of power P, with each face having same radius of curvature, is cut into two equal parts perpendicular to its principal axis. The power of one part of the lens will be: (A)2P (B) P (C) 4P (D) P/2​

Respuesta :

Otras preguntas

A double-convex lens of power P, with each face having same radius of curvature, is cut into two equal parts perpendicular to its principal axis. The power of one part of the lens will be: (A)2P (B) P (C) 4P (D) P/2