A) 5 mm
B) 10 mm
C) 15 mm
D) 20 mm
Correct Answer: C
Solution :
Here, angle subtended by moon \[=\frac{3500}{3\cdot 5\times {{10}^{5}}}radians\] Also, focal length of the mirror, \[f=R/2=1\cdot 5\,\,m\] \[=1.5\times {{10}^{3}}mm\] \[\therefore \] \[I=f\theta =1\cdot 5\times {{10}^{3}}\times \frac{3500}{3\cdot 5\times {{10}^{6}}}\] \[=15\,\,cm\]You need to login to perform this action.
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