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 \[\Rightarrow \] \[I=\frac{E}{R}=\frac{3}{30}=0.1A\]You need to login to perform this action.
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