In a Young's double slit experiment, the separation between the slits is 0.15 m m . In the experiment, a source of light of wavelength 589 n m is used and the interference pattern is observed on a screen kept 1.5 m  away. The separation between the successive bright fringes on the screen is:

Option 1 - <p><span class="mathml" contenteditable="false"> <math> <mn>4.9</mn> <mtext> </mtext> <mi mathvariant="normal">m</mi> <mi mathvariant="normal">m</mi> </math> </span></p>
Option 2 - <p><span class="mathml" contenteditable="false"> <math> <mn>6.9</mn> <mtext> </mtext> <mi mathvariant="normal">m</mi> <mi mathvariant="normal">m</mi> </math> </span></p>
Option 3 - <p><span class="mathml" contenteditable="false"> <math> <mn>3.9</mn> <mtext> </mtext> <mi mathvariant="normal">m</mi> <mi mathvariant="normal">m</mi> </math> </span></p>
Option 4 - <p><span class="mathml" contenteditable="false"> <math> <mn>5.9</mn> <mtext> </mtext> <mi mathvariant="normal">m</mi> <mi mathvariant="normal">m</mi> </math> </span></p>
6 Views|Posted 6 months ago
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1 Answer
A
6 months ago
Correct Option - 4
Detailed Solution:

  Energy     Volume   = M L 2 T - 2 L 3 = M L - 1 T - 2

 The distance between two successive bright fringes is fringe width ( β ) .

β = λ D d = 589 * 10 - 9 * 1.5 0.15 * 10 - 3 = 5.9 m m

 

 

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Physics Wave Optics 2025

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