Two coherent monochromatic point sources $S_{1}$ and $S_{2}$ of wavelength $\lambda=600 \mathrm{~nm}$ are placed symmetrically on either side of the center of the circle as shown. The sources are separated by a distance $d=1.8 \mathrm{~mm}$. This arrangement produces interference fringes visible as alternate bright and dark spots on the circumference of the circle. The angular separation between two consecutive bright spots is $\Delta \theta$. Which of the following options is/are correct?

(B) At $P_{2}$ the order of the fringe will be maximum; (C) The total number of fringes produced between $P_{1}$ and $P_{2}$ in the first quadrant is close to 3000

(A) A dark spot will be formed at the point $P_{2}$

(D) The angular separation between two consecutive bright spots decreases as we move from $P_{1}$ to $P_{2}$ along the first quadrant

JEE Advanced · 2017 · 21 May 2017 · Paper 2 · Wave Optics

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Two coherent monochromatic point sources

S_{1}

and

S_{2}

of wavelength

λ = 600 nm

are placed symmetrically on either side of the center of the circle as shown. The sources are separated by a distance

d = 1.8 mm

. This arrangement produces interference fringes visible as alternate bright and dark spots on the circumference of the circle. The angular separation between two consecutive bright spots is

Δ θ

. Which of the following options is/are correct?

(A)A dark spot will be formed at the point $P_{2}$
(B)At $P_{2}$ the order of the fringe will be maximum✓ Correct
(C)The total number of fringes produced between $P_{1}$ and $P_{2}$ in the first quadrant is close to 3000✓ Correct
(D)The angular separation between two consecutive bright spots decreases as we move from $P_{1}$ to $P_{2}$ along the first quadrant

Answer: (B), (C)

Solution

Path difference at point P on the circle is

Δ x = d cos θ

. At

P_{1}

(

θ = 90°

Δ x = 0

(maxima). At

P_{2}

(

θ = 0°

Δ x = d = nλ

, giving

n = \frac{1.8 \times 1 0 ^{- 3}}{600 \times 1 0 ^{- 9}} = 3000

(maxima, maximum order). The angular fringe width

∣ \frac{d θ}{d n} ∣ = \frac{λ}{d s i n θ}

increases as

θ

increases (decreases as we move from

P_{1}

P_{2}

is incorrect).

JEE Advanced · 2017 · 21 May 2017 · Paper 2 · One or more correct options

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JEE Advanced · 2017 · 21 May 2017 · Paper 2 · Wave Optics

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