Chapter 23, Problem 71P

To determine

The position and magnification of the final image when an object is placed is placed 96.5 cm from a glass lens (n=1.52)with one concave surface of radius 22 cm and one convex surface of radius 18.5 cm.

Answer to Problem 71P

Solution:

The image position is 169.75 cm from the lens in the object side and the magnification is 1.75

Explanation of Solution

Formula used:

lens maker’s formula $\frac{1}{f}=(1.52-1)\left(-\frac{1}{22}+\frac{1}{18.5}\right)$ where f is the focal length, r₁ is the radius of curvature of first surfce and r₂ is radius of curvature (RoC)of second surface of the lens. n is the refractive index of the lens material.

Lens equation, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

u and v are lens to object and lens to image distances repectively, negative when opposite to optical path and positive when in the direction of optical path.

Magnification, $m=\left|\frac{v}{u}\right|$

Calculation:

For first concave surface the sign of RoC is negative as it is opposite to optical path.

Thus, r₁=-22 cm

For second convex surface the sign of RoC is negative as it is opposite to optical path.

Thus, r₂=-18.5 cm

n=1.52

Thus, $\frac{1}{f}=(1.52-1)\left(-\frac{1}{22}+\frac{1}{18.5}\right)=\frac{91}{20350}$

u=-96.5 cm

Using lens equation,

$\frac{1}{v}+\frac{1}{96.5}=\frac{91}{20350}$

This gives, v=-169.75 cm

Magnification, m=169.75/96.5=1.76.