A person's near point is 50 , cm and his far | vedclass

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(B) For reading,the person needs to see an object at the standard near point $(u = -25 \, cm)$ as if it were at his own near point $(v = -50 \, cm)$.U

Vedclass · Accepted Answer

$2 \, D$ and $-0.33 \, D$

Vedclass · Answer

(B) For reading,the person needs to see an object at the standard near point $(u = -25 \, cm)$ as if it were at his own near point $(v = -50 \, cm)$.Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:$\frac{1}{f} = \frac{1}{-50} - \frac{1}{-25} = \frac{-1 + 2}{50} = \frac{1}{50} \, cm^{-1}$.$f = 50 \, cm = 0.5 \, m$.Power $P = \frac{1}{f(m)} = \frac{1}{0.5} = +2 \, D$.For seeing distant stars,the person needs to see an object at infinity $(u = \infty)$ as if it were at his far point $(v = -3 \, m)$.Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:$\frac{1}{f} = \frac{1}{-3} - \frac{1}{\infty} = -\frac{1}{3} \, m^{-1}$.Power $P = \frac{1}{f(m)} = -\frac{1}{3} \approx -0.33 \, D$.

$A$ person's near point is $50 \, cm$ and his far point is $3 \, m$. The power of the lenses he requires for $(i)$ reading and $(ii)$ seeing distant stars are:

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