Question

A near -sighted man can clearly see object only up-to a distance of 100 cm and beyond this. what is the number of the spectacles lens necessary for the remedy of this defect?​

Answer 1

Answer:

Maximum distance of distinct vision =400cm. So image of object at infinity is to be formed at 400cm

Use lens formula, v1−u1=f1

−4001−∞1=f1

P=−0.25D.

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Answer 2

Answer:

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

Here

$v = 2.5$

(Distance of retina as position of images is fixed )

$u = - x$

$\frac{1}{f} = \frac{1}{2.5} + \frac{1}{x}$

For f min : x is minimum,

$\frac{1}{ f_{min}} = \frac{1}{2.5} + \frac{1}{25}$

For f max : x is maximum ,

$\frac{1}{ f_{max} } = \frac{1}{2.5} + \frac{1}{ \infty }$

For near sighted man lens should make the image of the object within 100 cm range

For lens

$u = - \infty \\ v = - 100 \\ \frac{1}{ f_{lens} } = \frac{1}{ - 100} - \frac{1}{ - \infty }$