A) remain the same
B) be quadrupled
C) be doubled
D) be halved
Correct Answer: B
Solution :
Given: Initial distance between the slits \[{{d}_{1}}=d\] Final distance between the slits\[{{d}_{2}}=\frac{d}{2}\] Initial distance between slits and screen\[{{D}_{1}}=D\] Final distance between slit and screen\[{{D}_{2}}=2D\] The fringe width is given by \[\beta =\frac{\lambda D}{d}\propto \frac{D}{d}\] Hence, \[\frac{{{\beta }_{1}}}{{{\beta }_{2}}}=\frac{{{D}_{1}}}{{{d}_{1}}}\times \frac{{{d}_{2}}}{{{D}_{2}}}=\frac{D}{d}\times \frac{d/2}{2D}=\frac{1}{4}\] \[{{\beta }_{2}}=4{{\beta }_{1}}\] Hence, the fringe width will be quadrupled.
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