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Question

A person can read at the closest distance of 50cm. In order for him to read at 30cm. Identify the defect, nature, and power of the lens?


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Solution

Step 1: Type of defect and type of lens used for correction

  1. Myopia or nearsightedness is the defect here, it occurs when the eye losses its ability to focus on far-away objects.
  2. Objects near the eye are visible to patients with this defect
  3. The type of lens to be used here is a diverging lens to correct the defect

Step 2: Given

u= Object distance from the optic center

v= Image distance from the optic center

f= the focal length of the lens

u=-30cmv=-50cm

Step 3: Formula used

from the lens formula,

1f=1v-1u

Step 4: Calculation

1f=1v-1u1f=1-50+130f=-75cm

Power=100-75=-1.33D

Hence, the defect is myopia, the lens used is diverging and power is -1.33D.


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