Chapter 23, Problem 62P

To determine

The location and magnification of the final image formed by second lens when two 25 cm focal length converging lenses 16.5 cm apart with object placed at 35 cm before first lens.

Answer to Problem 62P

Solution:

The final image will be formed 18.49 mm after second lens with magnification of 0.65.

Explanation of Solution

Formula used:

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

f, u and v are lens to focal, 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:

Object position from first lens, $u_1=-35$ cm

Focal length of first and seconf lenses are f₁, f₂ respectively.

f₁ = f₂ = 25 mm

Image distance from first lens is v₁

$\frac{1}{v_1}+\frac{1}{35}=\frac{1}{f_1}$

This gives, $v_1=\text{ 87}.\text{5 cm}$

This image will act as virtual object for second lens. As the distance between two lenses is 16.5 cm

$u_2=\text{87}.\text{5}-\text{16}.\text{5}=\text{71 cm}$

Let position of final image with respect to second lens be v₂

Using lens equation,

$\frac{1}{v_2}-\frac{1}{71}=\frac{1}{f_2}$

This gives, v₂=18.49 cm

Total magnification, $m=\left|\frac{v_2{v}_1}{u_2{u}_1}\right|$=0.65