A far-sighted student has a near point of 1.0 m. Calculate the focal length (in cm) of the glasses needed so the near point will be normal (25 cm).
A) 72
B) 25
C) 33
D) 100
E) −33

Question

Quizplus · Accepted Answer

We use the formula for lens-maker's equation:

1/f = (n-1) * (1/R1 - 1/R2)

Where f is the focal length, n is the refractive index of the lens material (we assume it to be 1.5), R1 is the radius of curvature of the first surface of the lens, and R2 is the radius of curvature of the second surface.

Since we want the near point to be 25 cm, the lens should create a virtual image of the object at 25 cm from the eyes. This means the lens behaves like a magnifying glass.

Using the formula for magnification, we have:

m = -d/f

Where d is the distance of the object (which is the near point) before the lens. The minus sign indicates that the image is virtual, which matches our requirement.

Plugging in the given values, we have:

d = 1.0 m (since the near point is 1.0m)

m = -25 cm / 1.0 m = -0.25

Substituting these values in the magnification formula, we get:

-0.25 = -1.5 / f

Solving for f, we get f = -6 cm.

However, this focal length is negative which means we need a diverging lens to get the desired virtual image. We want a converging lens, so we take the absolute value of f:

|f| = 6 cm

But the answer choices are in cm, so we convert 6 cm to 60 mm.

Thus, the answer is: Choice C, 33 mm.

A Far-Sighted Student Has a near Point of 1