A person wears normal spectacles in which the | vedclass

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(B) For the spectacles:Let the far point of the eye be at a distance $x$ from the spectacles.The lens equation is $\frac{1}{v} - \frac{1}{u} = \frac{1

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$-4.54$

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(B) For the spectacles:Let the far point of the eye be at a distance $x$ from the spectacles.The lens equation is $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.For a distant object,$u = \infty$ and $v = -x \, cm$.$\frac{1}{-x} - \frac{1}{\infty} = \frac{1}{f} \implies f = -x \, cm = -\frac{x}{100} \, m$.Power $P = \frac{1}{f} = -\frac{100}{x} \, D$.Given $P = -5 \, D$,so $-5 = -\frac{100}{x} \implies x = 20 \, cm$.For contact lenses:The distance between the lens and the eye is $0$. The far point is at a distance $x + 2 \, cm$ from the eye (or the lens).So,the new object distance $u' = -(20 + 2) = -22 \, cm = -0.22 \, m$.To see distant objects $(u = \infty)$,the image must be formed at the far point $v' = -22 \, cm = -0.22 \, m$.Using the lens formula: $P' = \frac{1}{f'} = \frac{1}{v'} - \frac{1}{u'} = \frac{1}{-0.22} - \frac{1}{\infty} = -\frac{1}{0.22} \approx -4.54 \, D$.

$A$ person wears normal spectacles in which the distance between the glasses and the eyes is approximately $2 \, cm$. The power required is $-5 \, D$. If he wears contact lenses,then the required power is ......... $D$.

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