Lense- focal point, diagram position of image

[alicia113]

ok so here is the quesiton..

An object is loacted 3.5 cm from the optical centre of a lens. The lense produces an image of magnification that is +0.667 the size of the object

a) is this lens a diverging or converging lens? explain

b) Find the position of the image and the focal length of the lens using the magnification equation and the thin lens equation

c) make a diagram of the situation, showing the axis of symmetry and the prinicple axis of the lens, its optical centre itrs principle focal points and the positions of the object and image

Work

a) diverging lens (dont know why really but ill google it)

b) M=3.5/? = +0.667

So it would end up being 3.5/0.667 = 5.24? Am I correct?

and i can't get anything else please help!

 

[alicia113]

The lens is a diverging lens, because the image it creates is smaller in size than the object and is upright (indicated by positive magnification, which happens with a diverging lens. With a converging lens, this reduction in size would only happen when the image becomes inverted, giving a negative magnification.

The thin lens equation is 1/O + 1/I = 1/f

O - distance from of object
I - distance from image

f - focal length
We'll use that later.The magnification, M = -I/O

Therefore, 0.667 = -I/3.5

-I = 2.3345

I = -2.3345 cm

Distance of image from lens = 2.3345 cm

The image is 2.3345 cm in front of the lens, since diverging lenses' images are always in front of them.

As for the focal length, using the thin lens equation from before:

1/3 + -1/2.3345 = 1/f

Through algebraic manipulation, f = -10.52 cm

Therefore, the focal length of the lens is -10.52 cm.
Negative focal length confirms that the lens is a diverging lens.

MY UPDATED WORK IS IT CORRRECT?>!?

 

[alicia113]

FOR THE FOCAL LENGTH

i think i messed up...

should it be 1/3.5+ -1/2.3345 = 1/f

= -7.01

 

[ehild]

Your up-updated work is correct.

ehild