A machinist with normal vision has a near point at 25 cm. The machinist wears eyeglasses in order to do close work. The power of the lenses is +1.75 diopters. With these eyeglasses, what is the near point of the machinist?

Respuesta :

Answer:

17.4 cm

Explanation:

Power of lens = +1.75 diopters

Focal length of lens

[tex]f=\frac{100}{1.75}=\frac{400}{7}[/tex]

This is a convex lens as focal the diopter given is positive which makes the focal length positive. Image distance will be negative.

v = -25

[tex]\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\\\Rightarrow \frac{1}{-25}-\frac{1}{u}=\frac{1}{\frac{400}{7}}\\\Rightarrow -\frac{1}{u}=\frac{7}{400}+\frac{1}{25}\\\Rightarrow \frac{1}{u}=-\frac{7}{400}-\frac{1}{25}\\\Rightarrow \frac{1}{u}=-0.0575\\\Rightarrow u=-17.4\ cm[/tex]

∴ The new near point is 17.4 cm