d0400620
contestada


Example: What power of spectacle lens is needed to correct the vision of a nearsighted person
whose far point is 30.0 cm? Assume the spectacle (corrective) lens is held 1.50 cm away from the
eye by eyeglass frames.​

Respuesta :

Concave or diverging lens is needed correct the vision of a nearsighted person  whose far point is 30.0 cm

Solution:

Nearsighted person will not be able to see the distant objects clearly

Now, we want the near sighted person to see the distant objects clearly

The lens equation is given as:

[tex]P = \frac{1}{d_o} + \frac{1}{d_i}[/tex]

For distant vision, [tex]d_0 = \infty[/tex]

[tex]d_i = image\ distance\\\\d_0 = lens\ to\ retina\ distance[/tex]

The image of 30 cm from eye will be 28.5 cm to left of spectacle lens

Therefore,

[tex]d_i = -28.5\ cm = 0.285\ meter[/tex]

From lens equation,

[tex]P = \frac{1}{ \infty } + \frac{1}{-0.285\ meter}\\\\P = 0 - 3.51 \\\\P = -3.51\ diopter[/tex]

The negative power -3.51 D denotes that a concave or diverging lens is needed