Chapter 23, Problem 57P

To determine

(a)

The position, height and type of the image of $2.4cm$ high insect $1.3m$ away from a lens of focal length $135mm$.

Answer to Problem 57P

Solution:

The image will be formed $15.06cm$ after the lens and have a height of $0.28cm$. The image will be real.

Explanation of Solution

Given:

Focal length, $f=135mm=13.5cm$.

Object distance, $u=-1.3m=-130cm.$

Formula used:

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

Where, $f,u$ and $v$ are lens to focal, lens to object and lens to image distances respectively, negative when opposite to optical path and positive when in the direction of optical path.

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

Calculation:

From the given data we know that, $f=135mm=13.5cm$, $u=-1.3m=-130cm.$ Substituting them in lens equation we get-

$\frac{1}{v}+\frac{1}{130}=\frac{1}{13.5}$

This gives, $v=15.06cm$

Image height $=m\times$ object height =$\left|\frac{15.06}{130}\right|\times 2.4=0.28$ cm

This will be a real image.

To determine

(b)

The position, height and type of the image if focal length is $-135mm$

Answer to Problem 57P

Solution:

The virtual image will be formed $12.23cm$ before the lens and have a height of $0.22cm$. The image will be virtual.

Explanation of Solution

Given:

Focal length, $f=135mm=13.5cm$.

Object distance, $u=-1.3m=-130cm.$

Formula used:

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

Where, $f,u$ and $v$ are lens to focal, lens to object and lens to image distances respectively, negative when opposite to optical path and positive when in the direction of optical path.

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

Calculation:

$f=-135mm=-13.5cm$

$u=-1.3m=-130cm$

Using lens equation,

$\frac{1}{v}+\frac{1}{130}=-\frac{1}{13.5}$

This gives, $v=-12.23cm$

Image height $=m\times$ object height =$\left|\frac{12.23}{130}\right|\times 2.4=0.22$ cm

This will be a virtual image as image distance is negative.