Optics Problems

[sallyspears]

1. The problem statement, all variables and given/known data

When one mirror of a michelson interforemeter is translated by 0.0114cm, 523 fringes are observed. Calculate the wavelength of the light.


2. Relevant equations

wavelength = 2dm/m

3. The attempt at a solution

wavelength = 2x0.0114 / 523
= 4.39 ^10

------------------------------------------------------------------

1. The problem statement, all variables and given/known data

An object 2.5cm high is placed 12 cm in front of a thin lens focal length +3cm. How can i work out the image distance, and the magnification, and nature of the image.

2. Relevant equations

1/f = 1/v - 1/u

M = v/u

3. The attempt at a solution

Image distance i have worked out to be 4cm.

Magnification = v/u = 4/12 = 0.33?

I am unsure what a "nature of the image means"


Thanks for any help, is much appreciated

[lanedance]

try drawing a ray diagram to scale to check your calcs, easy & quick plus helps you understand what the equation is doing

the nature of the image usually consists of 2 things

inverted - upside down relative to object
real or virtual - reall means it is on the opposite side of the lens form the object

once again they will be easily apprent from your ray diagram & once you have done a few of these wioll have a feel for what the nature of the image will be

[Redbelly98]

Welcome to PF :smile:


wavelength = 2x0.0114 / 523

Looks good so far ...

= 4.39 ^10

Hmmm, looks like you entered something wrong into your calculator. Does it make sense for a wavelength of light to be such a large number? Also ... what are the units here???


1. The problem statement, all variables and given/known data

An object 2.5cm high is placed 12 cm in front of a thin lens focal length +3cm. How can i work out the image distance, and the magnification, and nature of the image.

2. Relevant equations

1/f = 1/v - 1/u

Should be
1/f = 1/v + 1/u
but it was probably a simple typo, as you got the correct answer of 4 cm.


M = v/u

Should be
M = -v/u

and a negative answer will indicate that the image is inverted.


3. The attempt at a solution

Image distance i have worked out to be 4cm.

Yes, good.

Magnification = v/u = 4/12 = 0.33?

Almost correct, just need the negative sign in the formula :smile:


I am unsure what a "nature of the image means"

They probably want to know if the image is real or virtual ... your text book or class notes should discuss what that means ...