Question

An object is 16.0cm to the left of a lens. The lens forms an image 36.0cm to the right of the lens.Part AWhat is the focal length of the lens?Part BIs the lens converging or diverging?Part CIf the object is 8.00mm tall, how tall is the image?Part DIs it erect or inverted?

Answer 1

A) 11.0 cm

The focal length of the lens can be find by using the lens equation:

`(1)/(f)=(1)/(p)+(1)/(q)`

where

p = 16.0 cm is the distance of the object from the lens

q = 36.0 cm is the distance of the image from the lens (positive, because it is on the opposite side to the lens with respect to the object)

Substituting into the equation, we find

`(1)/(f)=(1)/(16.0 cm)+(1)/(36.0 cm)=0.091 cm^(-1)`

`f=(1)/(0.091 cm^(-1))=11.0 cm`

B) Converging

The sign convention for the lenses states that:

- For a converging lens, the focal length is positive

- For a diverging lens, the focal length is negative

In this case, the focal length is positive (+11.0 cm), so the lenses must be a converging lens.

C) -18.0 mm

We can solve this part of the problem by using the magnification equation:

`(h_i)/(h_o)=-(q)/(p)`

where

`h_i` is the heigth of the image

`h_o=8.00 mm` is the height of the object

`q=36.0 cm` is the distance of the image from the lens

`p=16.0 cm` is the distance of the object from the lens

Solving the equation for
`h_i`, we find

`h_i = -h_o (q)/(p)=-(8.00 mm)(36.0 cm)/(16.0 cm)=-18.0 mm`

D) Inverted

For the sign convention, we have:

- if
`h_i` is positive, then the image is erect

- if
`h_i` is negative, then the image is upside-down (inverted)

In this case,
`h_i` is negative, therefore the image is inverted.