Question:
In a Young's double slit experiment, the separation between the slits is 0.15 mm. in the experiment, a source of light of wavelengh 589 nm 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 :
In a Young's double slit experiment, the separation between the slits is 0.15 mm. in the experiment, a source of light of wavelengh 589 nm 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 :
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Fringe width in YDSE: \(\beta=\dfrac{\lambda D}{d}\).
Updated On: Oct 31, 2026
- 4.9 mm
- 5.9 mm
- 6.9 mm
- 3.9 mm
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The Correct Option is B
Solution and Explanation
Step 1: \(\lambda=589\times10^{-9}\) m, \(D=1.5\) m, \(d=0.15\times10^{-3}\) m.
Step 2: \(\beta=\dfrac{\lambda D}{d}=\dfrac{589\times10^{-9}\times1.5}{0.15\times10^{-3}}=589\times10^{-9}\times10^{4}\) m.
Step 3: \(\beta=5.89\times10^{-3}\) m \(\approx5.9\) mm.
Answer: (B) 5.9 mm
Step 2: \(\beta=\dfrac{\lambda D}{d}=\dfrac{589\times10^{-9}\times1.5}{0.15\times10^{-3}}=589\times10^{-9}\times10^{4}\) m.
Step 3: \(\beta=5.89\times10^{-3}\) m \(\approx5.9\) mm.
Answer: (B) 5.9 mm
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