A) 16: 1
B) 1 : 16
C) 2 : 7
D) 7 : 2
Correct Answer: B
Solution :
The separation between source and photosensitive material at \[{{\phi }_{1}}=\] is 16m, therefore, intensity received by photosensitive material at \[=BA\] is \[{{\phi }_{2}}=\], where P is the power of source of light. At \[||\]s, the source is at (15, 0) and detector is at (19, 0), so the separation between them is 4 m. \[=0\] So, \[\therefore \,\,|\xi |\,=\,\left| \frac{-\Delta \phi }{\Delta t} \right|=\,\left| -\left( \frac{0-BA}{\frac{T}{4}-0} \right) \right|=\frac{4BA}{T}\]You need to login to perform this action.
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