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CBSE Class 10 Science Question 7 of 16

The Human Eye and the Colourful World — Question 7

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Question 7

Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect? Assume that the near point of the normal eye is 25 cm.

Answer

Hypermetropia can be corrected by using a convex lens as shown in the diagram below:

Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect? Assume that the near point of the normal eye is 25 cm. NCERT Class 10 Science CBSE Solutions.

An object at 25 cm forms an image at the near point of hypermetropic eye.

Given,

Near point of hypermetropic eye = 1 m =100 cm

Object distance, u = -25 cm

Image distance, v = -100 cm

According to the formula,

⇒1v−1u=1f⇒1−100−1−25=1f⇒−1+4100=1f⇒3100=1f⇒f=1003 cm⇒f=13 mPower=1f=113∴Power=3 D\phantom{\Rightarrow} \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f} \\[1 em] \Rightarrow \dfrac{1}{-100} - \dfrac{1}{-25} = \dfrac{1}{f} \\[1 em] \Rightarrow \dfrac{-1 + 4}{100} = \dfrac{1}{f} \\[1 em] \Rightarrow \dfrac{3}{100} = \dfrac{1}{f} \\[1 em] \Rightarrow f = \dfrac{100}{3} \text{ cm} \\[1 em] \Rightarrow f = \dfrac{1}{3} \text{ m} \\[1 em] \text{Power} = \dfrac{1}{f} = \dfrac{1}{\dfrac{1}{3}} \\[1 em] \therefore \text{Power} = 3 \text{ D}

Hence, the power of the lens required to correct this defect is 3 dioptre.