Myopia and Hypermetropia Numericals: Class 10 Science

Quick answer: A corrective lens forms a virtual image of the object at the point the defective eye can see. For myopia, take $u = -\infty$ and $v = -(\text{far point})$: a far point of 80 cm needs a concave lens of power −1.25 D. For hypermetropia, take $u = -25$ cm and $v = -(\text{near point})$: a near point of 1 m needs a convex lens of power +3.0 D. Use $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ and $P = \frac{1}{f}$ with $f$ in metres.

Near point and far point

The near point of 25 cm is also called the least distance of distinct vision.

How a corrective lens works

The lens does not focus light on the retina by itself. It makes a virtual image of the object at a distance the eye can already see clearly. The eye lens then focuses that image on the retina.

Both images are virtual and on the same side as the object, so $v$ is negative in both cases.

Numerical 1: myopia, far point 80 cm

The far point of a myopic person is 80 cm in front of the eye. Find the nature and power of the lens needed to see distant objects clearly.

Step 1: Write the values: $u = -\infty$, $v = -80$ cm.

Step 2: Use the lens formula:

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{-80} - 0 = -\frac{1}{80}$

So $f = -80$ cm $= -0.8$ m.

Step 3: Find the power:

$P = \frac{1}{f} = \frac{1}{-0.8} = -1.25 \text{ D}$

Answer: A concave (diverging) lens of focal length −80 cm and power −1.25 D.

Numerical 2: hypermetropia, near point 1 m

The near point of a hypermetropic eye is 1 m. Find the power of the lens needed to read at 25 cm.

Step 1: Write the values: $u = -25$ cm, $v = -100$ cm.

Step 2: Use the lens formula:

$\frac{1}{f} = \frac{1}{-100} - \frac{1}{-25} = \frac{-1 + 4}{100} = \frac{3}{100}$

So $f = \frac{100}{3}$ cm $\approx 33.3$ cm $= \frac{1}{3}$ m.

Step 3: Find the power:

$P = \frac{1}{f} = \frac{1}{1/3} = +3.0 \text{ D}$

Answer: A convex (converging) lens of focal length about +33.3 cm and power +3.0 D.

Quick shortcuts

Summary table

Exam tips

• Convert focal length to metres before finding power.
• State the nature of the lens from the sign: negative means concave, positive means convex.
• Write the values of $u$ and $v$ with signs before substituting.

Common mistakes

• Taking $u = -80$ cm in the myopia problem. The object is at infinity; 80 cm is where the image must form.
• Using $f$ in centimetres in $P = \frac{1}{f}$, which gives an answer 100 times too small.
• Making $v$ positive. The image formed by a spectacle lens here is virtual, so $v$ is negative.
• Mixing up which defect needs which lens.

Frequently Asked Questions

What lens corrects myopia?

A concave (diverging) lens of suitable negative power.

What is the power of the lens for a far point of 80 cm?

−1.25 D. The focal length is −80 cm, so P = 1/(−0.8 m) = −1.25 D.

What is the power of the lens for a near point of 1 m?

+3.0 D. With u = −25 cm and v = −100 cm, the focal length is 100/3 cm, which is 1/3 m.

Why is the image distance negative in these problems?

The corrective lens forms a virtual image on the same side as the object, and distances on that side are negative in the sign convention.

What is the least distance of distinct vision?

It is the near point of a normal eye, about 25 cm.