Question

an obiect 5cm high is placed 10cmin front of a converging lens offocal lenath 7cm. What is theimage distance, image height and magnification? Also, is it Uprightor Inverted, Real or Virtual, Enlarged or Reduced or Same Size

Answer 1

To determine the image distance we will use the following formula:

`(1)/(d_i)+(1)/(d_0)=(1)/(f)`

Where:

`\begin{gathered} d_i=\text{ distance of the image} \\ d_0=\text{ distance of the object} \\ f=\text{ focal distance} \end{gathered}`

We wll solve fo r the imae idistance. We will subtract both sides by 1/d0:

`(1)/(d_i)=(1)/(f)-(1)/(d_0)`

Now, we substitute the values:

`(1)/(d_i)=(1)/(7)-(1)/(10)`

Solving the operations:

`(1)/(d_i)=(3)/(70)`

Now we invert both sides:

`d_i=(70)/(3)=23.33cm`

Therefore, the distance of the image is 23.3 cm-.

To determine the image size w us e the followingdformula:

`(d_i)/(d_0)=-(h_i)/(h_0)`

Where:

`h_i,h_0=\text{ height of the image and height of the object}`

Now, we solve for the height of the image by multiplying both sides by the height of the object:

`-h_0(d_i)/(d_0)=h_i`

Now, we substitute the values:

`-(5cm)(23.33cm)/(10cm)=h_i`

Solving the operation:

`-11.66cm=h_i`

Therefore, the height ofthe image is -11.66 cm.

The magnification is given by:

`M=(h_i)/(h_0)`

Substituting we get:

`M=(23.33cm)/(10cm)=2.33`

Therefore, te magnification is 2.33

Since the image height i negative this means that the image isinvdeted.Since the magnification is greater than 1 this means that he image is enlargd. The iamgeimage is real.