A) ? 48, ? 40
B) ? 40, ? 48
C) ? 40, 48
D) ? 48, 40
Correct Answer: A
Solution :
When the final image is at the least distance of distinct vision, then \[m=-\frac{{{f}_{o}}}{{{f}_{e}}}\left( 1+\frac{{{f}_{e}}}{D} \right)=\frac{200}{5}\left( 1+\frac{5}{25} \right)=\frac{200\times 6}{5\times 5}=-\ 48\] When the final image is at infinity, then \[m=\frac{-{{f}_{o}}}{{{f}_{e}}}=\frac{200}{5}=-\ 40\]
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