How much must an uncorrected 1.00D myope accommodate to image the object which is at a distance of 33.33cm in retina?

Steps to Solve

1. Calculate the vergence of the object at 33.33 cm

The object is located at 33.33 cm. To find the vergence (L) of the light from the object entering the eye, we take the inverse of the distance in meters. $$ L = \frac{1}{0.3333 \text{ m}} \approx 3.00 \text{ D} $$

2. Account for the myopia

The person is a 1.00 D myope. This means their eye has 1.00 D too much power for distance vision (without accommodation). Since the person is uncorrected, the eye needs to reduce its power by 1.00 D to see far away. So, we can represent the eye's refractive error ($P_A$) as -1.00 D when thinking about how much accommodation is required.

3. Calculate the total power required

To focus the object at 33.33 cm, the eye needs a total power ($F_{fp}$) which can be calculated simply by summing two numbers obtained in the previous steps $$ F_{fp} = L + P_A = 3.00 \text{ D} - 1.00 \text{ D} = 2.00 \text{ D} $$

This means the eye needs to accommodate 2.00 D to see the object clearly.