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

 

[Nickie]

a) Based on the information given, we can determine that the lens is a converging lens. This is because the image produced is larger than the object, indicating that the light rays are being brought together to form the image.

b) Using the magnification equation, we can solve for the focal length of the lens:

M = -i/o = +0.667
i/o = -0.667
i = -0.667o

Using the thin lens equation, we can solve for the focal length:

1/f = 1/o + 1/i
1/f = 1/3.5 + 1/(-0.667o)
1/f = 1/3.5 - 1.5/o
f = 3.5/2.5 - 1.5o/2.5
f = 1.4 - 0.6o

Substituting the value of i from the magnification equation into this equation, we get:

f = 1.4 - 0.6(-0.667o)
f = 1.4 + 0.4o

Therefore, the focal length of the lens is 1.4 + 0.4o.

c) Please see the attached diagram for a visual representation of the situation described. The axis of symmetry is the line passing through the centre of the lens and perpendicular to its surface. The principal axis of the lens is the line passing through the centre of the lens and its principal focal points. The optical centre is the point where the axis of symmetry and the principal axis meet. The object is located 3.5 cm from the optical centre and the image is located at a distance of -0.667 times the object distance. The principal focal points are located 1.4 + 0.4o units away from the optical centre.

I hope this helps! Let me know if you have any further questions.