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  • a YDSE: D = 1 m, d = 1 mm and l = 5000 n m. The distance of 100th maxima from the central maxima is:Option: 1 <img alt="1/2 m" src="ht

a YDSE: D = 1 m, d = 1 mm and l = 5000 n m. The distance of 100th maxima from the central maxima is:

Option: 1

1/2 m


Option: 2

\sqrt 3 /2 m


Option: 3

1/\sqrt 3 m


Option: 4

does not exist 


Answers (1)

best_answer

As we learned

 

Fringe Width -

\beta = \frac{\lambda D}{d}
 

- wherein

\beta = y_{n+1}-y_{n}

y_{n+1}= Distance of\left ( n+1 \right )^{th}

Maxima= \left ( n+1 \right )\frac{\lambda D}{d}

y_{n}=Distance of n^{th}

 maxima = \frac{n\lambda D}{d}

 

 

for 100 _{th} max \\\\ d \sin \theta = 100 \lambda \\\\ \sin \theta = \frac{100 \times 5000\times 10 ^{-9}}1 \times 10 ^{-3}= 0.5 = 1/2 \\\\ y = D \tan \theta = 1 \times \tan 30 = 1/ \sqrt 3

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