Q19PE Repeat the previous problem for ... [FREE SOLUTION]

Chapter 26: Q19PE (page 956)

Repeat the previous problem for glasses that are 1.75 cm from the eyes.

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The far point for glasses that are 1.75 cm from the eyes is 26.8 cm.

Myopia is a vision condition in which visual pictures focus in front of the retina of the eye, creating vision problems of distant objects.

$d_{i} = 1.75 c m (\frac{1 m}{100 c m}) = 0.0175 m$

The power needed (P) to correct the vision can be obtained using the expression,

$P = \frac{1}{d_{0}} + \frac{1}{d_{i}}$

The power of the lens is equal to the reciprocal of the focal length. therefore,

$P = \frac{1}{f} P = \frac{1}{d_{0}} + \frac{1}{d_{i}}$

Rearrange and solve for the image's distance from the eye, we will get

$d_{i} = \frac{1}{P - \frac{1}{鈭�}} = \frac{1}{- 4 D - \frac{1}{鈭�}} d_{i} = - 0.25 m$

Hence, for the glasses that are 1.75 cm from the eyes, the distance between the eye and the image will be - 0.25 m - 0.0175 m = 0.268 m or 26.8 cm

Therefore, the far point for glasses that are 1.75 cm from the eyes is 26.8 cm

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