Question:
White light is used to illuminate the two slits in a Young's
double slit experiment. The separation between the slits is b
and the screen is at a distance d (>> b) from the slits. At a
point on the screen directly in front of one of the slits, certain
wavelengths are missing. Some of these missing
wavelengths are
White light is used to illuminate the two slits in a Young's
double slit experiment. The separation between the slits is b
and the screen is at a distance d (>> b) from the slits. At a
point on the screen directly in front of one of the slits, certain
wavelengths are missing. Some of these missing
wavelengths are
Updated On: Jun 14, 2022
- $\lambda=b^2/d$
- $\lambda=2b^2/d$
- $\lambda=b^2/3d$
- $\lambda=2b^2/3d$
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The Correct Option is C
Solution and Explanation
At P (directly infront of $S_1$ ) y = b / 2
$\therefore$ Path difference,
$\Delta X=S_2 P-S_1 P=\frac{y.(b)}{d}=\frac{\big(\frac{b}{2}\big)(b)}{d}=\frac{b^2}{2d}$
Those wavelengths will be missing for which
$\hspace25mm \Delta X=\frac{\lambda_1}{2},\frac{3 \lambda_2}{2},\frac{5 \lambda_3}{2}...$
$\therefore \hspace25mm \lambda_1=2 \Delta x=\frac{b^2}{d}$
$\hspace25mm \lambda_2=\frac{2 \Delta x}{3}=\frac{b^2}{3d}$
$\hspace25mm \lambda_3=\frac{2 \Delta x}{5}=\frac{b^2}{5d}$
$\therefore$ Correct options are (a) and (c).
$\therefore$ Path difference,
$\Delta X=S_2 P-S_1 P=\frac{y.(b)}{d}=\frac{\big(\frac{b}{2}\big)(b)}{d}=\frac{b^2}{2d}$
Those wavelengths will be missing for which
$\hspace25mm \Delta X=\frac{\lambda_1}{2},\frac{3 \lambda_2}{2},\frac{5 \lambda_3}{2}...$
$\therefore \hspace25mm \lambda_1=2 \Delta x=\frac{b^2}{d}$
$\hspace25mm \lambda_2=\frac{2 \Delta x}{3}=\frac{b^2}{3d}$
$\hspace25mm \lambda_3=\frac{2 \Delta x}{5}=\frac{b^2}{5d}$
$\therefore$ Correct options are (a) and (c).
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Concepts Used:
Young’s Double Slit Experiment
- Considering two waves interfering at point P, having different distances. Consider a monochromatic light source ‘S’ kept at a relevant distance from two slits namely S1 and S2. S is at equal distance from S1 and S2. SO, we can assume that S1 and S2 are two coherent sources derived from S.
- The light passes through these slits and falls on the screen that is kept at the distance D from both the slits S1 and S2. It is considered that d is the separation between both the slits. The S1 is opened, S2 is closed and the screen opposite to the S1 is closed, but the screen opposite to S2 is illuminating.
- Thus, an interference pattern takes place when both the slits S1 and S2 are open. When the slit separation ‘d ‘and the screen distance D are kept unchanged, to reach point P the light waves from slits S1 and S2 must travel at different distances. It implies that there is a path difference in the Young double-slit experiment between the two slits S1 and S2.
Read More: Young’s Double Slit Experiment