The least distance of distinct vision for a l | vedclass

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(B) For a long-sighted person,the near point is shifted to $x \ m$. To read a newspaper at $u = -x/2 \ m$,the lens must form a virtual image at the pe

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$1$

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(B) For a long-sighted person,the near point is shifted to $x \ m$. To read a newspaper at $u = -x/2 \ m$,the lens must form a virtual image at the person's near point,so $v = -x \ m$.Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.Substituting the values: $P = \frac{1}{f} = \frac{1}{-x} - \frac{1}{-x/2} = -\frac{1}{x} + \frac{2}{x} = \frac{1}{x}$.Given that the power $P = x \ D$,we have $x = \frac{1}{x}$.This implies $x^2 = 1$,so $x = 1$ (since distance must be positive).

The least distance of distinct vision for a long-sighted person is $x \ m$. He wants to read a newspaper placed at a distance $(x/2) \ m$. The power of the spectacle lens used by him is $x \ D$. What is the value of $x$?

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