A near sighted student has a near point of 0 1m... - JAMB Physics 2009 Question
A near sighted student has a near point of 0.1m and a focal length of 5.0cm. What is the student's far point?
A
0.200m
B
8.000m
C
0.125m
D
2.100m
correct option: e
Correct Answer: Option E
The person's image distance v will be at the distance of 0.1m = 10cm for him to see objects at a distance of u.F = 5.0cm
\( \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \implies \frac{1}{10} + \frac{1}{u} = \frac{1}{5} \\
\frac{1}{u} = \frac{1}{5} - \frac{1}{10} = \frac{2-1}{10} = \frac{1}{10} \\
\text{therefore } u = 10cm = 0.1m \)
\( \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \implies \frac{1}{10} + \frac{1}{u} = \frac{1}{5} \\
\frac{1}{u} = \frac{1}{5} - \frac{1}{10} = \frac{2-1}{10} = \frac{1}{10} \\
\text{therefore } u = 10cm = 0.1m \)
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