MCQ
The maximum number of possible interference maxima for slit-separation equal to twice the wavelength in Young’s double-slit experiment is
- AInfinite
- ✓$5$
- C$3$
- D$0$
✓
Answer
Correct option: B.
$5$
b
(b)For maxima $\Delta = d\sin \theta = n\lambda $
==> $2\lambda \sin \theta = n\lambda $==> $\sin \theta = \frac{n}{2}$
since value of sin $\theta$ can not be greater $1.$
$ n = 0, 1, 2$
Therefore only five maximas can be obtained on both side of the screen.
(b)For maxima $\Delta = d\sin \theta = n\lambda $
==> $2\lambda \sin \theta = n\lambda $==> $\sin \theta = \frac{n}{2}$
since value of sin $\theta$ can not be greater $1.$
$ n = 0, 1, 2$
Therefore only five maximas can be obtained on both side of the screen.
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