Question

A diffraction pattern is obtained using a beam of red light. Explain what happens if the red light is replaced by blue light.
A) Bands disappear.
B) No change.
C) The diffraction pattern becomes narrower and crowded.
D) The diffraction pattern becomes broader and further apart.

Answer 1

Hint:The diffraction pattern depends on the wavelength used. The red light has a longer wavelength than the blue light. So, the position and width of the bands will change accordingly. The position of the maxima or the minima is directly proportional to the wavelength of the light used.

Formula used:
-The position of the band is given by, ${y_n} = \dfrac{{n\lambda D}}{d}$ where $n$ is the order of the bands, $\lambda $ is the wavelength of the light used, $D$ is the distance from the screen to the slit and $d$ is the width of the slit.
-The angular width of the bands is given by the relation $d\sin {\theta _n} = n\lambda $ where $d$ is the width of the slit, $\theta $ is the angular width of the bands, $n$ is the order of the bands and $\lambda $ is the wavelength of the light used.

Complete step by step answer.
Step 1: Explain the difference in the position of the bands observed using the red light and the blue light.
The blue light has a shorter wavelength than the red light. If ${\lambda _r}$ is the wavelength of the red light and ${\lambda _b}$ is the wavelength of the blue light, then ${\lambda _b} < {\lambda _r}$ ------ (1)
The position of the band is given by, ${y_n} = \dfrac{{n\lambda D}}{d}$ --------- (2) where $n$ is the order of the bands, $\lambda $ is the wavelength of the light used, $D$ is the distance from the screen to the slit and $d$ is the width of the slit.
So, the positions of the bands will be at ${y_0} = 0$ , ${y_1} = \dfrac{{\lambda D}}{d}$ , ${y_2} = \dfrac{{2\lambda D}}{d}$ etc…
If we assume that the distance from the screen to the slit $D$ and the width of the slit $d$ remains the same in both scenarios (red light and blue light), then equation (2) shows that the position of the band depends only on the wavelength of the light used i.e., ${y_n} \propto n\lambda $ .
Now, since ${\lambda _b} < {\lambda _r}$ , when blue light is used, the bands will be crowded together.
Step 2: Explain the difference in the angular width of the bands observed using the red light and the blue light.
The angular width of the bands $\theta $ is given by the relation $d\sin {\theta _n} = n\lambda $ ------- (3).
Equation (3) suggests that if the width of the slit $d$ remains the same in both scenarios (red light and blue light) then the angular width of the bands will only depend on the wavelength i.e., $\sin {\theta _n} \propto n\lambda $
So, since ${\lambda _b} < {\lambda _r}$, when blue light is used, the bands will appear narrower.

Thus, the correct option is C.

Note: Here, the formula for the position of the bands is used under the assumption that the angular width $\theta $ of the bands is small. If $\theta $ is small, $\sin \theta \approx \theta $ and equation (3) can also be expressed as $d{\theta _n} = n\lambda $ . Maxima refers to a fringe of high intensity and minima refers to a fringe of zero intensity.