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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