Question
When the diffraction pattern from a certain slit illuminated with laser light $$\left( {\lambda = 6330\,\mathop {\text{A}}\limits^ \circ } \right)$$ is projected on a screen $$150\,cm$$ from the slit, the second minima on each side are separated by $$8\,cm.$$ This tells us that :
A.
the slit is approximately $$0.005\,cm$$ wide
B.
the slit is approximately $$0.05\,cm$$ wide
C.
$$\frac{a}{\lambda }$$ is approximately $$7.5$$ ($$a$$ is the slit width)
D.
$$\frac{a}{\lambda }$$ is approximately $$750$$
Answer :
the slit is approximately $$0.005\,cm$$ wide
Solution :
$$\eqalign{
& \sin {\theta _2} = \frac{{2\lambda }}{d}\,\,{\text{or}}\,\,{\theta _2} = {\sin ^{ - 1}}\left( {\frac{{2\lambda }}{d}} \right) \cr
& {\text{Given}}\,\,y = D\left( {2\,{\theta _2}} \right) = 8 \times {10^{ - 2}} \cr
& {\text{or}}\,\,1.5 \times 2 \times {\sin ^{ - 1}}\left( {\frac{{2\lambda }}{d}} \right) = 8 \times {10^{ - 2}} \cr
& \Rightarrow d = 0.005\,cm \cr} $$