1
$\begingroup$

The screen is far away. So we assume w/D << 1 where w is the distance between the slit and D the distance between screen and slit. But if this assumption bring about error which is comparable to the wavelength of light which is quite very small, wont the formulae become wrong. For example if we are using a red laser light, and the error due to our assumption while calculating the difference between the path length travelled by the light from both source comes around 300nm (around half the wavelength of red light) or any other multiple of this, than the anticipated bright fringe will become light and the light fringe dark. How can we make sure this isnt happening?

$\endgroup$
1
$\begingroup$

The exact path difference is:

$$ \Delta = \sqrt{D^2 + \big(\frac w 2\big)^2} -D $$

$$ \Delta = D\big[ \sqrt{1 + \big(\frac w {2D}\big)^2} - 1\big] $$

$$ \Delta = D\big[ \sqrt{1 + \epsilon^2} - 1\big]$$

with

$$ \epsilon \equiv \frac w {2D}$$

as a small parameter under $w/D \ll 1$.

Now Taylor expand around $\epsilon = 0$:

$$\Delta = D\big[ 1+\frac 1 2 \epsilon^2 -\frac 1 8 \epsilon^4 \big] $$

so the criterion for smallness is that the 2nd order term (in $\epsilon^4$) is less than some fraction of a wavelength:

$$ D \frac 1 8 \epsilon^4 =\frac D 8 \frac {w^4}{16D^4} \ll \lambda / n$$

Some ppl use $n=4$, some like $n=10$. Take your pick:

$$ \frac{w^4}{D^3\lambda} \ll \frac{128}n $$

For a 1 cm width and 300nm light, that means $D$ needs to be bigger than 20cm ish.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.