MCQ
What will be the angular width of central maxima in Fraunhoffer diffraction when light of wavelength $6000\,Å$ is used and slit width is $12 \times {10^{ - 5}}cm.$.......$ rad$
- A$2$
- B$3$
- ✓$1$
- D$8 $
✓
Answer
Correct option: C.
$1$
c
(c) Angular width $ = \frac{{2\lambda }}{d} = \frac{{2 \times 6000 \times {{10}^{ - 10}}}}{{12 \times {{10}^{ - 5}} \times {{10}^{ - 2}}}} = 1\,rad{\rm{.}}$
(c) Angular width $ = \frac{{2\lambda }}{d} = \frac{{2 \times 6000 \times {{10}^{ - 10}}}}{{12 \times {{10}^{ - 5}} \times {{10}^{ - 2}}}} = 1\,rad{\rm{.}}$
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