A) 1 m
B) 2 m
C) 3 m
D) 4 m
Correct Answer: B
Solution :
To obtain photographic print of same quality the light energy falling on unit area should be same ie, \[{{E}_{1}}\times {{t}_{1}}={{E}_{2}}\times {{t}_{2}}\] Now, \[{{E}_{1}}=\frac{{{I}_{1}}}{r_{1}^{2}}\] and \[{{E}_{2}}=\frac{{{I}_{2}}}{r_{2}^{2}}\] \[\therefore \] \[\frac{{{I}_{1}}}{r_{1}^{2}}\times {{t}_{1}}=\frac{{{I}_{2}}}{r_{2}^{2}}\times {{t}_{2}}\] \[\Rightarrow \] \[r_{2}^{2}=\frac{{{I}_{2}}}{{{I}_{1}}}\times \frac{{{I}_{2}}}{{{I}_{1}}}\times r_{1}^{2}\] \[=\frac{50}{75}\times \frac{12}{2}\times {{(1)}^{2}}=4\] \[{{r}_{2}}=2\,m\]
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