Diffraction patterns and lasers

[shyguy79]

Homework Statement

A laser with monochromatic light of frequency 6.33 × 10^14 Hz is fired through a diffraction grating.

If the first minimum in this diffraction pattern coincides with the position of the second order that was seen in the previous diffraction pattern, what is the width of the slit?

Homework Equations

$w=\frac{\gamma}{\sin\theta}$

$n\gamma=d\sin\theta_{n}$

The Attempt at a Solution

Here are my results from a previous question to which it refers

Using the following values: $\gamma=633nm$
and $\theta=39.3^{o}$
and the equation $n\gamma=dsin\theta_{n}$

Rearranged for distance between the slits $d=\frac{n\gamma}{sin\theta_{n}}$
then $d=\tfrac{5*633x10^{-9}m}{sin(39.3)}=5\mu m$ or 5.00E^{-6}m

Now to answer this question I've calculated the distance between the slits:

First calculate the distance between the slits second order using the equation $d=\frac{n\gamma}{sin\theta} = \frac{2*633E^{-9}m}{sin(39.3)}=2\mu m$ or 2.0E{}^{-6}
m.

This is where I don't know where to go... I know I need w... but how? Please help, I've had a thought:
by combining the two equations and substituting for $\sin\theta$ then maybe $\frac{n\gamma}{d}=\frac{\gamma}{w}$ so $w=\frac{\gamma d}{n\gamma}=\frac{d}{n}$

 

[Redbelly98]

Hello,

I'm getting confused in trying to follow what you did, so let's start by clearing some things up.

(I'll not worry that you're using $\gamma$ for wavelength when λ is the conventional symbol used by maybe 99.9% of all physics teachers and textbooks )

Homework Statement

A laser with monochromatic light of frequency 6.33 × 10^14 Hz is fired through a diffraction grating.

If the first minimum in this diffraction pattern coincides with the position of the second order that was seen in the previous diffraction pattern, what is the width of the slit?

Okay, for starters, diffraction gratings do not produce well-defined minima. Since you mention "the width of the slit", I am wondering if the current problem is actually for a single slit, whereas the previous problem did use a diffraction grating. Can you confirm this, or clarify what the setup really is for the two problems?

Homework Equations

$w=\frac{\gamma}{\sin\theta}$

That is either the 1st order minimum for a single slit of width w -- or possibly the 1st order maximum of a diffraction grating (or double slit) of grating (slit) spacing w.

$n\gamma=d\sin\theta_{n}$

Looks good. That gives the n^th-order maxima of a diffraction grating (or double slit) of grating (slit) spacing d.

The Attempt at a Solution

Here are my results from a previous question to which it refers

Using the following values: $\gamma=633nm$
and $\theta=39.3^{o}$
and the equation $n\gamma=dsin\theta_{n}$

Rearranged for distance between the slits $d=\frac{n\gamma}{sin\theta_{n}}$
then $d=\tfrac{5*633x10^{-9}m}{sin(39.3)}=5\mu m$ or 5.00E^{-6}m

It looks like the problem was to find d for a diffraction grating, if the 5th-order maximum makes an angle of 39.3° for light of wavelength 633 nm. Can you confirm my guess?

Now to answer this question I've calculated the distance between the slits:

First calculate the distance between the slits second order using the equation $d=\frac{n\gamma}{sin\theta} = \frac{2*633E^{-9}m}{sin(39.3)}=2\mu m$ or 2.0E{}^{-6} m.

What slits? If it's the same diffraction grating, the grating spacing doesn't change and is still 5.00 μm. Or is it really supposed to be some other grating?

This is where I don't know where to go... I know I need w... but how? Please help, I've had a thought:
by combining the two equations and substituting for $\sin\theta$ then maybe $\frac{n\gamma}{d}=\frac{\gamma}{w}$ so $w=\frac{\gamma d}{n\gamma}=\frac{d}{n}$