question_answer

Light of wavelength 589.3 nm is incident normally on a slit of width 0.1 mm. The angular width of the central diffraction maximum ac a distance of 1 m from the slit, is:

A) 0.68°          

B)                        0.34°     

C)                        2.05°                     

D)        none of these

Correct Answer: A

Solution :

                Let D be the distance between source and screen d the distance between coherent source then for central diffraction maxima, We have \[{{\lambda }_{sun}}\times {{T}_{sun}}={{\lambda }_{star}}\times {{T}_{star}}\] where \[\lambda \] is wavelength. Given, \[\Rightarrow \] \[\frac{{{T}_{sun}}}{{{T}_{star}}}=\frac{{{\lambda }_{star}}}{{{\lambda }_{sun}}}=\frac{350}{510}=0.68\] \[R={{R}_{1}}+{{R}_{2}}=10\Omega +20\Omega =30\Omega \]  \[E={{E}_{1}}-{{E}_{2}}=5V-2V=3V\] Angular width \[\Rightarrow \]                 \[I=\frac{E}{R}=\frac{3}{30}=0.1A\]