Solve Optics Problems: Calculate Wavelength & Image Distance, Magnification

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SUMMARY

The discussion focuses on solving optics problems involving a Michelson interferometer and a thin lens. The wavelength of light is calculated using the formula wavelength = 2dm/m, resulting in an approximate value of 4.39 x 10^-7 m after correcting calculation errors. Additionally, the image distance for an object placed in front of a thin lens with a focal length of +3 cm is determined to be 4 cm, with magnification calculated as M = -v/u, indicating the image is inverted. The nature of the image is clarified as either real or virtual, which can be confirmed through ray diagrams.

PREREQUISITES
  • Understanding of basic optics principles, including lens formulas.
  • Familiarity with the Michelson interferometer and its applications.
  • Knowledge of magnification calculations in optics.
  • Ability to interpret ray diagrams for image analysis.
NEXT STEPS
  • Study the principles of the Michelson interferometer and its wavelength calculations.
  • Learn about ray diagrams and their role in determining the nature of images formed by lenses.
  • Explore the concept of real vs. virtual images in optics.
  • Investigate common mistakes in optics calculations and how to avoid them.
USEFUL FOR

Students studying optics, physics educators, and anyone involved in experimental physics or optical engineering will benefit from this discussion.

sallyspears
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Homework Statement



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


Homework Equations



wavelength = 2dm/m

The Attempt at a Solution



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

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

Homework Statement



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.

Homework Equations



1/f = 1/v - 1/u

M = v/u

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
 
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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
 
Welcome to PF :smile:

sallyspears said:
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?

Homework Statement



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.

Homework 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.

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 textbook or class notes should discuss what that means ...
 

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