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# 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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## Step 1: Type of defect and type of lens used for correction Myopia or nearsightedness is the defect here, it occurs when the eye losses its ability to focus on far-away objects.Objects near the eye are visible to patients with this defectThe 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$u=-30cm\phantom{\rule{0ex}{0ex}}v=-50cm$Step 3: Formula usedfrom the lens formula,$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$Step 4: Calculation$⇒\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{f}=\frac{1}{-50}+\frac{1}{30}\phantom{\rule{0ex}{0ex}}⇒f=-75cm$Power$=\frac{100}{-75}=-1.33D$Hence, the defect is myopia, the lens used is diverging and power is $-1.33D$.  Suggest Corrections  0      Similar questions  Explore more