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प्रश्न
The near point of the eye of a person is 50 cm. Find the nature and power of the corrective lens required by the person to enable him to see clearly the objects placed at 25 cm from the eye?
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उत्तर
Given:
Object distance (u) = −25cm
Image distance (v) = −50cm
Focal length, f = ?
Using the lens formula
`1/"f" = 1/"v" - 1/"u"`
`1/"f" = 1/(-50) - 1/(-25)`
`1/"f" = -1/50 + 1/25`
`1/"f" = (-1+2)/50`
`1/"f" = 1/50`
`"f"` = 50 cm
Since the focal length of the lens is positive, the corrective lens required is a convex lens.
`"f"` = 50cm
`"f"` = 0.5m
Power (P) of the corrective lens will be
`"P" = 1/("f"("m"))`
`"P" = 1/0.5`
`"P" = + 2 "D"`
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संबंधित प्रश्न
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Given below is a diagram depicting a defect of the human eye. Study the same and answer the question that follow:
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(i) Label the parts 1 to 5 of the diagram.
(ii) State the function of the parts labelled 4 and 5.
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Given alongside is a diagram depicting a defect of the human eye. Study the same and answer the questions that follow:
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- Draw a labelled diagram to show how the above mentioned defect is rectified using the lens named above.
