Question
A myopic person has been using spectacles of power –1.0 dioptre for distant vision. During old age he also needs to use separate reading glass of power + 2.0 dioptres. Explain what may have happened.
✓
Answer
Power of spectacles, P = -1 D
$\therefore$ Focal length, f = -100 cm
That is, the far point of the person is at 100 cm.
Near point of the eye might have been normal (i.e., 25 cm).
The objects at infinity produce virtual images at 100 cm (using spectacles).
To see objects between 25 cm to 100 cm, the person uses the ability of accommodation of his eye-lens. This ability is partially lost in old age.
The near point of the eye may recede to 50 cm. He has, therefore, to use glasses of suitable power for reading.
Here,
Object distance, u = -25 cm
Image distance, v = -50 cm
Since, $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}=-\frac{1}{50}+\frac{1}{25}$
i.e., focal length, f = 50 cm
power $\text{P}=\frac{100}{\text{f}}=\frac{100}{50}=+2$ dipotre.
$\therefore$ Focal length, f = -100 cm
That is, the far point of the person is at 100 cm.
Near point of the eye might have been normal (i.e., 25 cm).
The objects at infinity produce virtual images at 100 cm (using spectacles).
To see objects between 25 cm to 100 cm, the person uses the ability of accommodation of his eye-lens. This ability is partially lost in old age.
The near point of the eye may recede to 50 cm. He has, therefore, to use glasses of suitable power for reading.
Here,
Object distance, u = -25 cm
Image distance, v = -50 cm
Since, $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}=-\frac{1}{50}+\frac{1}{25}$
i.e., focal length, f = 50 cm
power $\text{P}=\frac{100}{\text{f}}=\frac{100}{50}=+2$ dipotre.
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