Question

Suppose the distance from the lens system of the eye (cornea + lens) to the retina is \(18 \mathrm{mm}\). (a) What must the power of the lens be when looking at distant objects? (b) What must the power of the lens be when looking at an object \(20.0 \mathrm{cm}\) from the eye? (c) Suppose that the eye is farsighted; the person cannot see clearly objects that are closer than $1.0 \mathrm{m}$. Find the power of the contact lens you would prescribe so that objects as close as \(20.0 \mathrm{cm}\) can be seen clearly.

Short answer

Answer: The power of the contact lens required for a farsighted person to see clearly at 20 cm is 3.2 diopters (D).

Step-by-step solution

(a) Lens power for distant objects

First, we need to find the lens power when looking at distant objects. We can use the lens formula: \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \) As we are looking at distant objects, the object distance (d_o) is extremely large, so \( \frac{1}{d_o} \) is nearly equal to 0. Therefore, our equation simplifies to: \( \frac{1}{f} = \frac{1}{d_i} \) The distance to the retina (d_i) is 18 mm or 0.018 m. Now we calculate the focal length (f): \( f = d_i = 0.018m \) To find the required power (P) in diopters, we use the power formula: \( P = \frac{1}{f} \) Calculating the lens power when looking at distant objects: \( P = \frac{1}{0.018} = 55.5 \, D \)

(b) Lens power for close objects

Next, we need to find the lens power when looking at an object 20 cm (0.2 m) from the eye. We'll use the lens formula again: \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \) In this case, the object distance (d_o) is 0.2 m, and the image distance (d_i) remains 0.018 m. First, we need to find the new focal length (f). We can rearrange the lens formula for f: \( f = \frac{1}{ \frac{1}{d_o} + \frac{1}{d_i} } \) Plugging in the values: \( f = \frac{1}{ \frac{1}{0.2} + \frac{1}{0.018} } = 0.0167 \, m \) Now, we'll use the power formula again to find the new power (P') required when looking at close objects: \( P' = \frac{1}{f} \) Calculating the lens power when looking at close objects: \( P' = \frac{1}{0.0167} = 60 \, D \)

(c) Contact lens power for farsighted person

Since the person is farsighted and cannot see clearly objects closer than 1.0 m, we first need to find the power of their eyes (P_e) when looking at objects within that limit. Let's use the lens formula again with d_o = 1.0m and d_i = 0.018m: \( f_e = \frac{1}{ \frac{1}{1} + \frac{1}{0.018} } = 0.0176 \, m \) Now, we'll calculate the power of their eyes (P_e) using the power formula: \( P_e = \frac{1}{f_e} = 56.8 \, D \) Next, we need to find the power of the contact lens (P_c) required to see clearly at 20 cm. We already know from part (b) that the total power needed is 60 D. The total power will be the sum of the power of the eye (P_e) and the power of the contact lens (P_c). Therefore: \( 60 = 56.8 + P_c \) Solving for the power of the contact lens: \( P_c = 60 - 56.8 = 3.2 \, D \)