At the larger angle shown in part (c), the path lengths differ by
$3\text{\lambda}\text{/}2$ for rays from the top and bottom of the slit. One ray travels a distance
$\lambda $ different from the ray from the bottom and arrives in phase, interfering constructively. Two rays, each from slightly above those two, also add constructively. Most rays from the slit have another ray to interfere with constructively, and a maximum in intensity occurs at this angle. However, not all rays interfere constructively for this situation, so the maximum is not as intense as the central maximum. Finally, in part (d), the angle shown is large enough to produce a second minimum. As seen in the figure, the difference in path length for rays from either side of the slit is
D sin
$\theta $ , and we see that a destructive minimum is obtained when this distance is an integral multiple of the wavelength.
Thus, to obtain
destructive interference for a single slit ,
where
D is the slit width,
$\text{\lambda}$ is the light’s wavelength,
$\theta $ is the angle relative to the original direction of the light, and
m is the order of the minimum.
[link] shows a graph of intensity for single-slit interference, and it is apparent that the maxima on either side of the central maximum are much less intense and not as wide. This effect is explored in
Double-Slit Diffraction .
Calculating single-slit diffraction
Visible light of wavelength 550 nm falls on a single slit and produces its second diffraction minimum at an angle of
$45.0\text{\xb0}$ relative to the incident direction of the light, as in
[link] . (a) What is the width of the slit? (b) At what angle is the first minimum produced?
Strategy
From the given information, and assuming the screen is far away from the slit, we can use the equation
$D\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta =m\lambda $ first to find
D , and again to find the angle for the first minimum
${\theta}_{1}.$
Solution
We are given that
$\text{\lambda}=550\phantom{\rule{0.2em}{0ex}}\text{nm}$ ,
$m=2$ , and
${\theta}_{2}=45.0\text{\xb0}$ . Solving the equation
$D\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta =m\lambda $ for
D and substituting known values gives
Solving the equation
$D\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta =m\lambda $ for
$\text{sin}\phantom{\rule{0.2em}{0ex}}{\theta}_{1}$ and substituting the known values gives
We see that the slit is narrow (it is only a few times greater than the wavelength of light). This is consistent with the fact that light must interact with an object comparable in size to its wavelength in order to exhibit significant wave effects such as this single-slit diffraction pattern. We also see that the central maximum extends
$20.7\text{\xb0}$ on either side of the original beam, for a width of about
$41\text{\xb0}$ . The angle between the first and second minima is only about
$24\text{\xb0}$$\left(45.0\text{\xb0}-20.7\text{\xb0}\right)$ . Thus, the second maximum is only about half as wide as the central maximum.
As refractive index depend on other factors also but if we supply heat on any system or media its refractive index decrease.
i.e. it is inversely proportional to the heat.
ganesh
you are correct
Priyojit
law of multiple
Wahid
if we heated the ice then the refractive index be change from natural water