SATHEE: Optics 6 Question 50

Optics 6 Question 50

52. A coherent parallel beam of microwaves of wavelength $λ = 0.5 m m$ falls on a Young’s double slit apparatus. The separation between the slits is $1.0 m m$ . The intensity of microwaves is

measured on a screen placed parallel to the plane of the slits at a distance of $1.0 m$ from it as shown in the figure.

$(1998, 8$ M)

(a) If the incident beam falls normally on the double slit apparatus, find the $y$ -coordinates of all the interference minima on the screen.

(b) If the incident beam makes an angle of $30^{\circ}$ with the $x$ -axis (as in the dotted arrow shown in figure), find the $y$ -coordinates of the first minima on either side of the central maximum.

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Answer:

Correct Answer: 52. (a) $\pm 0.26 m, \pm 1.13 m$ (b) $0.26 m, 1.13 m$

Solution:

Given, $λ = 0.5 m m, d = 1.0 m m, D = 1 m$

(a) When the incident beam falls normally :

Path difference between the two rays $S_{2} P$ and $S_{1} P$ is

$Δ x = S_{2} P - S_{1} P \approx d \sin θ$

For minimum intensity,

$d \sin θ = (2 n - 1) \frac{λ}{2}$ , where $n = 1, 2, 3, \dots$

$\begin{aligned} \sin θ & = \frac{(2 n - 1) λ}{2 d} \\ = \frac{(2 n - 1) 0.5}{2 \times 1.0} = \frac{2 n - 1}{4} \end{aligned}$

As $\sin θ \leq 1$ therefore $\frac{(2 n - 1)}{4} \leq 1$ or $n \leq 2.5$

So, $n$ can be either 1 or 2 .

When $n = 1, \sin θ_{1} = \frac{1}{4}$ or $\tan θ_{1} = \frac{1}{\sqrt{15}}$

$n = 2, \sin θ_{2} = \frac{3}{4}$

or $\tan θ_{2} = \frac{3}{\sqrt{7}}$

$∴ y = D \tan θ = \tan θ$

$(D = 1 m)$

So, the position of minima will be

$\begin{aligned} y_{1} = \tan θ_{1} = \frac{1}{\sqrt{15}} m = 0.26 m \\ y_{2} = \tan θ_{2} = \frac{3}{\sqrt{7}} m = 1.13 m \end{aligned}$

And as minima can be on either side of centre $O$ .

Therefore there will be four minimas at positions $\pm 0.26 m$ and $\pm 1.13 m$ on the screen.

(b) When $α = 30^{\circ}$ , path difference between the rays before reaching $S_{1}$ and $S_{2}$ is

$Δ x_{1} = d \sin α = (1.0) \sin 30^{\circ} = 0.5 m m = λ$

So, there is already a path difference of $λ$ between the rays.

Position of central maximum Central maximum is defined as a point where net path difference is zero. So,

$\begin{aligned} Δ x_{1} = Δ x_{2} \\ or d \sin α = d \sin θ \\ or θ = α = 30^{\circ} \\ or \tan θ = \frac{1}{\sqrt{3}} = \frac{y_{0}}{D} \\ \begin{aligned} ∴ y_{0} & = \frac{1}{\sqrt{3}} m (D = 1 m) \\ y_{0} & = 0.58 m \end{aligned} \end{aligned}$

At point $P, Δ x_{1} = Δ x_{2}$

Above point $P Δ x_{2} > Δ x_{1}$ and

Below point $P Δ x_{1} > Δ x_{2}$

Now, let $P_{1}$ and $P_{2}$ be the minimas on either side of central maxima. Then, for $P_{2}$

$\begin{aligned} Δ x_{2} - Δ x_{1} & = \frac{λ}{2} \\ Δ x_{2} & = Δ x_{1} + \frac{λ}{2} = λ + \frac{λ}{2} = \frac{3 λ}{2} \end{aligned}$

$\begin{array}{ll} or & d \sin θ_{2} = \frac{3 λ}{2} \\ or & \sin θ_{2} = \frac{3 λ}{2 d} = \frac{(3) (0.5)}{(2) (1.0)} = \frac{3}{4} \\ ∴ & \tan θ_{2} = \frac{3}{\sqrt{7}} = \frac{y_{2}}{D} \\ or & y_{2} = \frac{3}{\sqrt{7}} = 1.13 m \end{array}$

Similarly by for $P_{1}$

$\begin{aligned} Δ x_{1} - Δ x_{2} = \frac{λ}{2} or Δ x_{2} = Δ x_{1} - \frac{λ}{2} = λ - \frac{λ}{2} = \frac{λ}{2} \\ or d \sin θ_{1} = \frac{λ}{2} \\ or \sin θ_{1} = \frac{λ}{2 d} = \frac{(0.5)}{(2) (1.0)} = \frac{1}{4} \\ ∴ \tan θ_{1} = \frac{1}{\sqrt{15}} = \frac{y_{1}}{D} \\ or y_{1} = \frac{1}{\sqrt{15}} m = 0.26 m \end{aligned}$