A person cannot see objects beyond 3 , m. Wha | vedclass

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Question

(C) The person is suffering from myopia (nearsightedness). The far point of the person is $x = 3 \, m$. The person wants to see objects at $y = 12 \,

Vedclass · Accepted Answer

$-1/4 \, D$

Vedclass · Answer

(C) The person is suffering from myopia (nearsightedness). The far point of the person is $x = 3 \, m$. The person wants to see objects at $y = 12 \, m$.To correct this,the lens must form a virtual image of an object placed at $y = 12 \, m$ at the person's far point $x = 3 \, m$.Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.Here,$v = -3 \, m$ and $u = -12 \, m$.$\frac{1}{f} = \frac{1}{-3} - \frac{1}{-12} = -\frac{1}{3} + \frac{1}{12} = \frac{-4 + 1}{12} = -\frac{3}{12} = -\frac{1}{4} \, m^{-1}$.Therefore,the power of the lens is $P = \frac{1}{f} = -0.25 \, D$.

$A$ person cannot see objects beyond $3 \, m$. What should be the focal length of the lens required to see objects at $12 \, m$?

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