Question

An object with a height of −0.060 m points below the principal axis (it is inverted) and is 0.120 m in front of a diverging lens. The focal length of the lens is −0.28 m. (Include the sign of the value in your answers.) What is the image distance? What is the magnification? What is the image height?

Answer 1

Explanation:

Given that,

Height of object = -0.060 m

Distance of object = 0.120 m

Focal length = -0.28 m

We need to calculate the image distance

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$

Where, f = focal length

v = image distance

u = object distance

Put the value into the formula

$\dfrac{1}{-0.28}=\dfrac{1}{v}+\dfrac{1}{-0.120}$

$\dfrac{1}{v}=\dfrac{1}{-0.28}-\dfrac{1}{0.120}$

$\dfrac{1}{v}=-\dfrac{250}{21}$

$v=-0.084\ m$

Negative sign shows the image on the same side of the object and the image is virtual.

(b). We need to calculate the magnification

Using formula of magnification

$m= \dfrac{-v}{u}$

$m=\dfrac{0.084}{0.120}$

$m=0.7$

Positive value of magnification shows the object and image is inverted.

(c). We need to calculate the image height

Using formula of magnification

$m=\dfrac{h'}{h}$

$0.7=\dfrac{h'}{-0.060}$

$h'=0.7\times(-0.060)$

$h'=-0.042\ m$

The height of the image is 0.042 m.

Hence, This is the required solution.