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Physics · 2022 · 3 marks
CBSE 2022 · Region 5 · Set 1 · Q7
How can you differentiate whether a pattern is produced by a single slit or double slits? Derive the expression for the angular position of (i)…
How can you differentiate whether a pattern is produced by a single slit or double slits? Derive the expression for the angular position of (i) bright and
(ii)
dark fringes produced in a single slit diffraction.
Marking-scheme solution
Difference in the pattern of fringes due to single slit and double slits $\displaystyle 1$ Derivation of angular position of (i) Bright fringe and
(ii)
Dark fringe $\displaystyle 1$+$\displaystyle 1$
In the pattern of fringes produced by a single slit, the central fringe (band) is brighter as compared to other fringes i.e intensity goes on decreasing as the order (n) of the maxima increases, while in the fringe pattern produced by double slits all bright fringes including central fringe are of same intensity.
Alternatively:
In the fringe pattern produced by single slit, the fringe at the centre is wider as compared to the width of other bright fringes, while in the fringe pattern produced by double slits all bright fringes are of equal width.
Calculation of angular position
For the slit of width 'a ' and angle of diffraction ' $\displaystyle \theta$ '
Path difference $\displaystyle (\Delta \mathrm{p})=\mathrm{a} \sin \theta$
(i)
Condition for Bright Fringe, $\displaystyle \Delta \mathrm{p}=(2 \mathrm{n}+1) \frac{\lambda}{2}$
$\displaystyle \therefore \mathrm{a} \sin \theta=(2 \mathrm{n}+1) \frac{\lambda}{2}$
$$\sin \theta=\frac{(2 n+1) \lambda}{2 a} \quad(\mathrm{n}=$\displaystyle 1,2$ \ldots)
$$for small angle, $\displaystyle \sin \theta \approx \theta$
$$\theta=\frac{2 n+1) \lambda}{2 a}
$$(ii) Condition for dark fringe , $\displaystyle \Delta \mathrm{p}=\mathrm{n} \lambda$
$$\begin{aligned}
& \mathrm{a} \sin \theta=\mathrm{n} \lambda \quad(\mathrm{n}=$\displaystyle 1,2$ \ldots)
& \therefore \sin \theta=\frac{n \lambda}{2 a}
\end{aligned}
$$for small angle, $\displaystyle \sin \theta \approx \theta$
$$\theta=\frac{n \lambda}{a}
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