A) \[7200\overset{\text{o}}{\mathop{\text{A}}}\,\]
B) \[6800\overset{\text{o}}{\mathop{\text{A}}}\,\]
C) \[5500\overset{\text{o}}{\mathop{\text{A}}}\,\]
D) \[4000\overset{\text{o}}{\mathop{\text{A}}}\,\]
Correct Answer: D
Solution :
Here, \[n=3,\,{{y}_{n}}=1.2mm=1.2\times {{10}^{-3}}m\] \[D=1m,\,d=1\times {{10}^{-3}}m\] For the position of maxima \[{{y}_{n}}=\frac{nD\lambda }{d}\] or \[\lambda =\frac{{{y}_{n}}d}{nD}\] So, the wavelength is given by \[\lambda =\frac{{{y}_{n}}d}{nD}=\frac{1.2\times {{10}^{-3}}\times 1\times {{10}^{-3}}}{3\times 1}\] or \[\lambda =0.4\times {{10}^{-6}}m=4\times {{10}^{-7}}m=4000\overset{\text{o}}{\mathop{\text{A}}}\,\]
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