Question
The first diffraction minimum due to the single slit diffraction is seen at $$\theta = {30^ \circ }$$ for a light of wavelength $$5000\,\mathop {\text{A}}\limits^ \circ $$ falling perpendicularly on the slit. The width of the slit is
A.
$$2.5 \times {10^{ - 5}}cm$$
B.
$$1.25 \times {10^{ - 5}}cm$$
C.
$$10 \times {10^{ - 5}}cm$$
D.
$$5 \times {10^{ - 5}}cm$$
Answer :
$$10 \times {10^{ - 5}}cm$$
Solution :
For first minimum,
$$\eqalign{
& d\sin \theta = \lambda \cr
& \Rightarrow d = \frac{\lambda }{{\sin \theta }} = \frac{{5000 \times {{10}^{ - 8}}}}{{\sin {{30}^ \circ }}} \cr
& = \frac{{5000 \times {{10}^{ - 8}}}}{{\frac{1}{2}}} = 10 \times {10^{ - 5}}cm \cr} $$