Chapter 23, Problem 65P

To determine

(a) To Determine:

A ray diagram and estimate the location and the relative size of the final image

Answer to Problem 65P

Solution:

Explanation of Solution

Given:

A lighted candle is placed $36cm$ in front of a converging lens of focal length $f_1=13cm$, which in $56cm$ in front of another converging lens of focal length $f_2=16cm$.

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:

Ray diagram logic of the location and the relative size of the final image when a lighted candle is placed $36cm$ in front of a converging lens of focal length $f_1=13cm$, which in $56cm$ in front of another converging lens of focal length $f_2=16cm$ is shown below-

Here we can observe that image is real and diminished.

To determine

(b)

The value of position and relative size of the final image.

Answer to Problem 65P

Solution:

Final image will be formed $29.03cm$ at the right of 2^nd lens with magnification of $0.46$.

Explanation of Solution

Given:

A lighted candle is placed $36cm$ in front of a converging lens of focal length $f_1=13cm$, which in $56cm$ in front of another converging lens of focal length $f_2=16cm$.

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:

Positions of the candle and its first image with respect to first lens are $u_1=36cm$ and $f_1=13cm$.Using lens equation,

$\begin{array}{l}\frac{1}{v_1}+\frac{1}{36}=\frac{1}{13}\\ \frac{1}{v_1}=\frac{1}{13}-\frac{1}{36}\end{array}$

This gives, $v_1=20.35cm$

Position of the first image with respect to second lens is $u_2=-(56cm-v_1)=-35.65cm$

Considering $v_2$ to be the position of the final image with respect to second lens,

$\begin{array}{l}\frac{1}{v_2}+\frac{1}{35.65}=\frac{1}{16}\\ \frac{1}{v_2}=\frac{1}{16}-\frac{1}{35.65}\end{array}$

This gives, $v_2=29.03cm$

The total magnification i.e. the relative size of the candle is

$m=\left|\frac{v_1{v}_2}{u_1{u}_2}\right|$$=0.46$.