a person cannot see objects nearer than 75 cm from his eyes, normal vision is 25 cm from the eye. find the nature, focal length and power of lens used for defective vision
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the defective vision is hypermetropia and the formula used to find this is 1/f=1/v-1/u
given u=25 v=75
1/f=1/75-1/25=
2/75
given u=25 v=75
1/f=1/75-1/25=
2/75
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Answer:
The defect of the vision is hypermetropia.
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