Question
A beam of light of $$\lambda = 600\,nm$$ from a distant source falls on a single slit $$1\,mm$$ wide and the resulting diffraction pattern is observed on a screen $$2\,m$$ away. The distance between first dark fringes on either side of the central bright fringe is
A.
$$1.2\,cm$$
B.
$$1.2\,mm$$
C.
$$2.4\,cm$$
D.
$$2.4\,mm$$
Answer :
$$2.4\,mm$$
Solution :
According to question diagram is shown as
$$\tan \theta = \frac{{\frac{x}{2}}}{2}$$
For small $$\theta $$ and when $$\theta $$ is counted in rad, $$\tan \theta \approx \theta $$
$$\eqalign{ & {\text{So,}}\,\,\theta = \frac{{\frac{x}{2}}}{2} \cr & \Rightarrow \frac{\lambda }{a} = \frac{x}{4} \cr & \Rightarrow x = \frac{{4\lambda }}{a} = \frac{{4 \times 600 \times {{10}^{ - 9}}}}{{{{10}^{ - 3}}}} \cr & = 24 \times {10^{ - 4}}m \cr & = 2.4 \times {10^{ - 3}}m \cr & = 2.4\,mm \cr} $$
According to question diagram is shown as
$$\tan \theta = \frac{{\frac{x}{2}}}{2}$$
For small $$\theta $$ and when $$\theta $$ is counted in rad, $$\tan \theta \approx \theta $$
$$\eqalign{ & {\text{So,}}\,\,\theta = \frac{{\frac{x}{2}}}{2} \cr & \Rightarrow \frac{\lambda }{a} = \frac{x}{4} \cr & \Rightarrow x = \frac{{4\lambda }}{a} = \frac{{4 \times 600 \times {{10}^{ - 9}}}}{{{{10}^{ - 3}}}} \cr & = 24 \times {10^{ - 4}}m \cr & = 2.4 \times {10^{ - 3}}m \cr & = 2.4\,mm \cr} $$
