A) 137 cm
B) 0.683
C) 2.07 cm
D) 1.98 cm
Correct Answer: A
Solution :
Given, \[\lambda =560\,nm=560\times {{10}^{-9}}m\] \[d=0.20\,\,mm=2\times {{10}^{-4}}m\] and \[D=2\,m\] As the radius of image disc \[R=1.22\frac{\lambda D}{b}\] \[=\frac{1.22\times 560\times {{10}^{-9}}\times 2}{2\times {{10}^{-4}}}\] \[=6.832\times {{10}^{-3}}m\] \[=0.683\,cm\] So, the radius of central bright fring formed on the wall \[=2R=2\times 0.683=1.37\,cm\]
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