A) 0.576
B) 0.348
C) 0.192
D) 0.096
Correct Answer: C
Solution :
[c] : Distance between the first minimum on the left and the first minimum on the right is also the width of central maximum. Width of central maximum,\[W=\frac{2\lambda D}{a}\]where, X = Wavelength of light a = Width of the slit D = Distance of the screen from the slit \[\therefore \]\[a=\frac{2\lambda D}{W}\] Here, \[\lambda =6000\overset{\text{o}}{\mathop{\text{A}}}\,=6000\times {{10}^{-10}}m,\] \[D=80cm=80\times {{10}^{-}}^{2}m\] \[V=5mm=5\times {{10}^{-3}}m\] \[\therefore \]\[a=\frac{2\times 6000\times {{10}^{-10}}m\times 80\times {{10}^{-2}}m}{5\times {{10}^{-3}}m}\] \[=19.2\times {{10}^{-5}}m=0.192\times {{10}^{-3}}m=0.192mm\]
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