A person with a defective eye vision ia unable to see nearer objects more than 1.5m. He wants to read a book at 30 cm . Find the nature, focal length and power of lens.
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This person suffers from the defect of hypermetropia.
so,
u = -30cm and v = -1.5m = -150cm
Using lens formula
1/f = 1/v - 1/u
1/f = 1/-150 - 1/-30
1/f = 1/-150 + 1/30
1/f = -1+5/150
1/f = 4/150
f = +37.5cm
so, the positive sign show that the lens was needed a convex lens..
nature is virtual and erect
Power (P) = 1/f = 1/37.5cm
= 100/37.5(in m)
= 2.67D
Hope this helps you...
so,
u = -30cm and v = -1.5m = -150cm
Using lens formula
1/f = 1/v - 1/u
1/f = 1/-150 - 1/-30
1/f = 1/-150 + 1/30
1/f = -1+5/150
1/f = 4/150
f = +37.5cm
so, the positive sign show that the lens was needed a convex lens..
nature is virtual and erect
Power (P) = 1/f = 1/37.5cm
= 100/37.5(in m)
= 2.67D
Hope this helps you...
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