Everyday SciencePhysics
The near point of a hypermetropic eye is 1m. The power of the lens required to correct this defect is
1D
2D
3D
4D
3D
In the problem, it is given that the near point of defective eye is 1 m and that of a normal eye is 25 cm. Hence u=โ25 cm. The lens used forms its virtual image at near point of hypermetropic eye i.e., v=โ1m=โ100 cm. Using lens formula, we have :- v1โโu1โ=f1โโ1001โโโ251โ=f1โf=3100โ=0.33mpower=f(inmetres)1โ=+3.0D
3D In the problem, it is given that the near point of defective eye is 1 m and that of a normal eye is 25 cm. Hence u=โ25 cm. The lens used forms its virtual image at near point of hypermetropic eye i.e., v=โ1m=โ100 cm. Using lens formula, we have :- v1โโu1โ=f1โโ1001โโโ251โ=f1โf=3100โ=0.33mpower=f(inmetres)1โ=+3.0D
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