Find the width of an opening that light is scattered by - given the angle to the 2nd interference maximum and wavelength

[JoeyBob]

Homework Statement:

See attached

Relevant Equations:

angle=wavelength/(2a)

The destructive interference equation for small angles is angle=wavelength/(2a), where a is the width. I assume it means destructive interference since its talking about areas where no light is present.

Using the equation after changing degrees into radians I get the answer of 2491 nm when the answer shoould be 9982 nm. the answer is approx. 4 times as large. Where am I going wrong here?

 

[BvU]

Hi,

Study http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/sinslit.html to find a factor of 2 (basically: you have the wrong relevant equation) and consider that the exercise text (were you not to be bothered retyping it ?) talks about the second angle to find another factor of 2 ...

$\$

 

[JoeyBob]

Hi,

Study http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/sinslit.html to find a factor of 2 (basically: you have the wrong relevant equation) and consider that the exercise text (were you not to be bothered retyping it ?) talks about the second angle to find another factor of 2 ...

$\$

i see, I was using a double slit eqn instead of a single slit. Looking at your link, tanx=x=y/D.

So i don't know what y is nor d.

d=y/x=y/0.108 = 9.2593y

Now I can use y=(m*wavelength*D)/a to find width. m i assume is 2 because second angle measurement...

0.108d=(2*539 nm *d)/a

a=9981.4815

This is good. Thanks. I have more understanding over slits now that *hopefully* I won't just haphazardly use an equation that seems appropriate.