The far point of a myopic person is 80, cm in | vedclass

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Question

(A) For a myopic person,the far point is at a finite distance $v = -80\, cm$.To see objects at infinity $(u = \infty)$ clearly,the lens must form a vi

Vedclass · Accepted Answer

$-1.25\, D$

Vedclass · Answer

(A) For a myopic person,the far point is at a finite distance $v = -80\, cm$.To see objects at infinity $(u = \infty)$ clearly,the lens must form a virtual image at the person's far point.Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.Substituting the values: $\frac{1}{f} = \frac{1}{-80} - \frac{1}{\infty} = \frac{1}{-80} - 0 = -\frac{1}{80}\, cm^{-1}$.Thus,the focal length $f = -80\, cm = -0.8\, m$.The power of the lens is given by $P = \frac{1}{f(m)}$.$P = \frac{1}{-0.8} = -1.25\, D$.

$(I)$ Presbyopia	$(A)$ Sphero-cylindrical lens
$(II)$ Hypermetropia	$(B)$ Convex lens of proper power may be used close to the eye
$(III)$ Astigmatism	$(C)$ Concave lens of suitable focal length
$(IV)$ Myopia	$(D)$ Bifocal lens of suitable focal length

The far point of a myopic person is $80\, cm$ in front of the eye. What is the power of the lens required to enable him to see very distant objects clearly?

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