Question

A converging lens of focal length 8.000 cm is 19.5 cm to the left of a diverging lens of focal length -6.00 cm. A coin is placed 12.1 cm to the left of the converging lens.

(a) Find the location of the coin's final image relative to the diverging lens. (Include the sign of each answer. Enter a negative value if the image is to the left of the diverging lens.)
_______________ cm

(b) Find the magnification of the coin's final image.

Answer 1

`(1)/(d_(o))+(1)/(d_(i))=(1)/(f)\\d_(i)=(d_(o)f)/(f-d_(o))=((-4.1cm)(-6cm))/(-6cm-(-4.1cm)))=-12.94cm\\`

`M=M(d_(i))/(d_(o))=(-1.95)((-2.94cm))/(-4.1cm)=-1.39cm`

Step-by-step explanation:

Firs of all, we calculate the position of the image of the ring with the converging lens. After this we take the image of the ring as the object for the diverging lens (we have to use the lens' equation, and we assume that objects to the left are positive and images to the right are positive).

(a) Hence, we have for the first step

`(1)/(d_(o))+(1)/(d_(i))=(1)/(f)\\(1)/(d_(i))=(1)/(f)-(1)/(d_(o))\\d_(i)=(d_(o)f)/(d_(o)-f)=(12.1cm*8cm)/(12.1cm-8cm)=23.6cm\\`

so, the image is at a distance of 23.6 cm to the right of the converging lens. The distance between both converging and diverging lens is 19.5cm. Hence the position of the image of the ring (produced by the first lens) is

23.6cm - 19.5cm = 4.1cm

4.1 cm to the right of the diverging lens. Hence we have for a diverging lens

`(1)/(d_(o))+(1)/(d_(i))=(1)/(f)\\d_(i)=(d_(o)f)/(f-d_(o))=((-4.1cm)(-6cm))/(-6cm-(-4.1cm)))=-12.94cm\\`

where we take do=-4.1cm beacuse the object is to the right. The image is to the left of the diverging lens

(b)

The magnification is

`M=(-d_(i))/(d_(o))=(-23.6cm)/(12.1cm)=-1.95`

this constant is a factor for the second magnification

`M=M(d_(i))/(d_(o))=(-1.95)((-2.94cm))/(-4.1cm)=-1.39cm`

I hope this is useful for you

regards