A) 625 nm, 500 nm
B) 380 nm, 525 nm
C) 380 nm, 500 nm
D) 400 nm, 500 nm
Correct Answer: A
Solution :
For maxima, \[d\,\sin \theta =n\lambda \] |
\[\lambda =\frac{d\theta }{n}\](\[\because \sin \theta \approx \theta \]when\[\theta \]is small) |
\[=\frac{2500}{n}nm\] |
Take\[n=4,5\]for\[\lambda =625nm\]and\[500nm.\] |
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