In an interference arrangement similar to Young’s double slit experiment, but having four slits S1, S2, S3 and S4 are illuminated with coherent light wave of wavelength 400 nm. Sources are synchronized in way such that S1, S2 and S4 have zero phase difference but S3 is out of phase with others. In central wide part of screen intensity due to all individual slits is I0. The consecutive slits are separated by a distance d = 1 mm. Distance between plane of slits and screen is D = 10 mPoint O is a point on screen lie on perpendicular bisector of S1S2. ‘P’ is the point on the screen closest to point ‘O’ having intensity 2I0 when only S1 and S2 are illuminated. Now if all slits are illuminated, what will be new intensity at point ‘P’?

Let distance OP = yPath difference of S1 and S3 at P=ydD=λ4y=λD4d…..(i)⇒y=dNow path difference of S1 and S3=3d2(2d)D=3d2D=3λ4And path difference of S1 and S4 is =2dD(3d)=32λPhasor diagramAR=2A IR=4I0

In an interference arrangement similar to Young’s double slit experiment, but having four slits S1, S2, S3 and S4 are illuminated with coherent light wave of wavelength 400 nm. Sources are synchronized in way such that S1, S2 and S4 have zero phase difference but S3 is out of phase with others. In central wide part of screen intensity due to all individual slits is I0. The consecutive slits are separated by a distance d = 1 mm. Distance between plane of slits and screen is D = 10 mPoint O is a point on screen lie on perpendicular bisector of S1S2. ‘P’ is the point on the screen closest to point ‘O’ having intensity 2I0 when only S1 and S2 are illuminated. Now if all slits are illuminated, what will be new intensity at point ‘P’?

Let distance OP = yPath difference of S1 and S3 at P=ydD=λ4y=λD4d…..(i)⇒y=dNow path difference of S1 and S3=3d2(2d)D=3d2D=3λ4And path difference of S1 and S4 is =2dD(3d)=32λPhasor diagramAR=2A IR=4I0

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

In an interference arrangement similar to Young’s double slit experiment, but having four slits S₁, S₂, S₃ and S₄ are illuminated with coherent light wave of wavelength 400 nm. Sources are synchronized in way such that S₁, S₂ and S₄ have zero phase difference but S₃ is out of phase with others. In central wide part of screen intensity due to all individual slits is I₀. The consecutive slits are separated by a distance d = 1 mm. Distance between plane of slits and screen is D = 10 m
Point O is a point on screen lie on perpendicular bisector of S₁S₂. ‘P’ is the point on the screen closest to point ‘O’ having intensity 2I₀ when only S₁ and S₂ are illuminated. Now if all slits are illuminated, what will be new intensity at point ‘P’?

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

4I₀

2I₀

3I₀

Zero

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Let distance OP = y
Path difference of S₁ and S₃ at $P = \frac{yd}{D} = \frac{λ}{4}$
$y = \frac{λD}{4 d} \dots . . (i) \Rightarrow y = d$
Now path difference of S₁ and S₃
$= \frac{3 \frac{d}{2} (2 d)}{D} = \frac{3 d^{2}}{D} = \frac{3 λ}{4}$
And path difference of S₁ and S₄ is $= \frac{2 d}{D} (3 d) = \frac{3}{2} λ$
Phasor diagram