MCQ
The maximum number of possible interference maxima for slit separation equal to $1.8\,\lambda $, where $\lambda $ is the wavelength of light used, in a Young's double slit experiment is
- Azero
- ✓$3$
- Cinfinite
- D$5$
✓
Answer
Correct option: B.
$3$
b
As $\sin \theta=\frac{n \lambda}{d}$ and $\sin \theta$ cannot be $ \ne \,1$
As $\sin \theta=\frac{n \lambda}{d}$ and $\sin \theta$ cannot be $ \ne \,1$
$\therefore 1=\frac{n \lambda}{1.8 \lambda}$
or $n=1.8$
Hence maximum number of possible interference maximas, $0, \pm 1$ i.e. $3$
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