## wave-optics Question 25

Question: Q. 1. (i) Explain the two features to distinguish between the interference pattern in Young’s double slit experiment with the diffraction pattern obtained due to single slit.

(ii) A monochromatic light of wavelength $500 \mathrm{~nm}$ is incident normally on a single slit of width $0.2 \mathrm{~mm}$ to produce a diffraction pattern. Find the angular width of the central maximum obtained on the screen.

Estimate the number of fringes obtained in Young’s double slit experiment with fringe width $0.5 \mathrm{~mm}$, which can be accommodated within the region of total angular spread of the central maximum due to single slit.

U] [Delhi I, II 2017]

## Show Answer

Solution:

Ans. (i) Try yourself, Similar to Q. 1, Short Answer Type Questions-I

(ii) Angular width of central maximum

$$ \begin{aligned} \omega & =\frac{2 \lambda}{a} \ & =\frac{2 \times 500 \times 10^{-9}}{0.2 \times 10^{-3}} \text { radian } \mathbf{1} \ & =5 \times 10^{-3} \text { radian } \end{aligned} $$

Now,

$$ \begin{equation*} \beta=\frac{\lambda D}{d} \tag{1} \end{equation*} $$

Linear width of central maxima in the diffraction pattern

$$ \beta^{\prime}=\frac{2 \lambda D}{d} $$

Let ’ $n$ ’ be the number of interference fringes which can be accommodated in the central maxima.

$$ \begin{aligned} \therefore \quad n \times \beta & =\beta^{\prime} \ n & =\frac{2 \lambda D}{d} \times \frac{d}{\lambda D} \ n & =2 \end{aligned} $$

[Award the last $1 / 2$ mark if the student writes the answer as 2 (taking $d=a$ ), or just attempts]

[CBSE Marking Scheme 2017]

AII Q. 2. In Young’s double slit experiment, deduce the condition for (a) constructive, and (b) destructive interference at a point on the screen. Draw a graph showing variation of intensity in the interference pattern against position ’ $x$ ’ on the screen. (ii) Compare the interference pattern observed in Young’s double slit experiment with single slit diffraction pattern, pointing out three distinguishing features.

U] [Delhi 2016]

Ans. (i)

From figure,

Path difference $=\left(S_{2} P-S_{1} P\right)$

$\left(S_{2} P\right)^{2}-\left(S_{1} P\right)^{2}=\left[D^{2}+\left(x+\frac{d}{2}\right)^{2}\right]-\left[D^{2}+\left(x-\frac{d}{2}\right)^{2}\right]$

$\left(S_{2} P+S_{1} P\right)\left(S_{2} P-S_{1} P\right)=2 x d$

$\Rightarrow \quad S_{2} P-S_{1} P=\frac{2 x d}{\left(S_{2} P+S_{1} P\right)}$

For $\quad x, d«D$

$$ S_{2} P-S_{1} P=\frac{2 x d}{2 D}=\frac{x d}{D} $$

For constructive interference,

$\begin{aligned} S_{2} P-S_{1} P & =n \lambda, \ \Rightarrow \quad \frac{x d}{D} & =n \lambda\end{aligned}$

$$ x=\frac{n \lambda D}{d} $$

$\therefore$ Position of the $n^{\text {th }}$ bright fringe on screen :

For destructive interference

(ii) (a) The interference pattern has number of equally spaced bright and dark bands, while in the diffraction pattern the width of the central maximum is twice the width of other maxima. $1 / 2$

(b) In Interference all bright fringes are of equal intensity, whereas in the diffraction pattern the intensity falls as order of maxima increases. $1 / 2$

(c) In Interference pattern, maxima occurs at an angle $\frac{\lambda}{a}$, where, $a$ is the slit width, whereas in diffraction pattern, at the same angle i.e., at $\frac{\lambda}{a}$ first minima occurs. (Here ’ $a$ ’ is the size of the slit)

(Any other distinguishing feature)

[CBSE Marking Scheme 2016]