A satisfactory photographic print is obtained when the exposure time is 10 sec at a distance of 2 m from a 60 cd lamp. The time of exposure required for the same quality print at a distance of 4 m from a 120 cd lamp is
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a
5 sec
b
10 sec
c
15 sec
d
20 sec
answer is D.
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Detailed Solution
I1D12t1=I2D22t2 Here D is constant and I=Lr2 So L1r12×t1=L2r22×t2⇒60(2)2×10=120(4)2×t⇒20sec