Question

A 1.5-cm object is placed 0.50 m to the left of a diverging lens with a focal length of 0.20 m. A converging lens with a focal length of 0.17 m is located 0.08 m to the right of the diverging lens. What is the height and orientation with respect to the original object of the final image?

Answer 1

The object for the converging lens is upright and 0.429 cm tall, the image of this converging lens is inverted and 1.375 cm high

Step-by-step explanation:

Let

`d_(o)=distance of object[tex]\\f=focal length\\d_(i)=distance of image\\I_(h)=Image height`

For diverging lens:

`d_(o) = 0.50\\f = -0.20\\(1)/(d_(o))+(1)/(-0.20)\\(1)/(d_(i))=(1)/(-0.20)-(1)/(0.50)=-7\\d_(o)=-(1)/(7)`

Magnification =
`(d_(i))/(d_(o))= -(1)/(7)÷ 0.5 = -0.286`

Image height
`= -0.286 * 1.5 = -0.429 cm` (negative sign means the image is virtual, inverted.

This image is
`(1)/(7)` meter to left of the center of the diverging lens.

The converging lens is located 0.08 m to the right of the diverging lens

The distance between the image of the diverging lens and center of the converging lens =
`(1)/(7) + 0.08 = 0.229 m`

The image of the diverging lens becomes the object of the converging lens.

`d_(o) = 0.223\\f = 0.17\\(1)/(d_(i))=(1)/(0.17)-(1)/(0.223)=0.715\\d_(i)=0.715m to the right of the converging lens`

`Magnification =(d_(i))/(d_(o)) = (0.715)/(0.223)=3.206\\image height=3.206 * 0.429 = 1.375 cm.`