An eye condition is corrected using a +4.0D lens - AQA - A-Level Physics - Question 1 - 2022 - Paper 5

To calculate the magnification (M) produced by a lens, we can use the formula:

$M = \frac{v}{u}$

Where:

• $v$ is the image distance (in meters)
• $u$ is the object distance (in meters)

For a lens of power $P = +4.0D$:

$v = \frac{1}{P} = \frac{1}{4} = 0.25\text{ m} = 25\text{ cm}$

The object distance $u$ is given as 75 cm, or 0.75 m. Thus, we calculate:

$M = \frac{25}{75} = \frac{1}{3}$

Thus, the magnification produced by the lens is approximately $0.33$, indicating that the image is one-third the size of the object.

The name of X is cornea.

The primary optical function of X (the cornea) is to refract light as it enters the eye.

The name of Y is lens.

The primary optical function of Y (the lens) is to accommodate and focus light onto the retina.

The name of Z is pupil.

The primary optical function of Z (the pupil) is to control the amount of light entering the eye.