Lesson 5 – The simple magnifying glass

• Define the terms far point and near point for the unaided eye (For the normal eye, the far point may be assumed to be at infinity and the near point is conventionally taken as being a point 25 cm from the eye).

• Define angular magnification• Derive an expression for the angular

magnification of a simple magnifying glass for an image formed at the near point and at infinity.

Magnifying glass

Imagine a lens of focal length 5cm, with an object of height 1cm placed a distance of 3.5 cm from the lens

Magnifying glass

The lines to not intersect so there is no real image.

Magnifying glass

However if the rays are extended backwards they do intersect to form a virtual image

Magnifying glass

The eye is tricked into believing that the rays originated at the point where the extended rays intersect. This is where the image is.

Magnifying glass

The image is 11.7cm from the lens and has a height of 3.3cm. It appears magnified.

Magnifying glass

Using 1/f = 1/u + 1/v1/5 = 1/3.5 + 1/v1/v = 1/5 – 1/3.5 = -0.0857v = -11.67cm (virtual), M = -v/u = -(-11.67)/3.5 = 3.3

The Human Eye

The lens of your eye can be pulled or squashed to change how thick it is and thus enable it to focus on objects at different distances.

Near point

The near point is defined as the closest point on which an unaided human eye can focus without straining. How far? Let’s try!

Near point

The near point is defined as the closest point on which an unaided human eye can focus without straining. It is taken to be 25cm for a normal eye, but can change greatly with age.

Far point

The far point is defined as the largest distance an eye can focus without straining. It is taken as infinity for the normal eye. In practice infinity means anything larger than a few metres! (rays arriving almost parallel)

Short sightedness (Myopia)

For a normal eye...

But if the eyeball is too long or the lens too powerful...

So the short sighted person’s far point is nearer than infinity...

Q. How can this be corrected?

Apparent size/angle subtended

Apparent size/angle subtended

Coin

Apparent size/angle subtended

Coin

Angle subtended at near point

tanθ ≈ h/25 = θWithout a magnifying glass the closest the object can be focused is at the near point. It will appear normal size:

θh

25 cm

Object viewed through a lens (object closer to the lens than the

focal point)

25 cm

u

f

θihi

h

If a virtual image (v is negative) is formed at 25cm and the distance of the object from the lens is u then

1/f = 1/u + 1/-25u = 25f/(25 + f)

25 cm

u

f

θihi

h

Let θi be the angle the image subtends at the lens. From simple geometry we obtain θi = hi/25 cm = h/u = h(25 + f)/25f

25 cm

u

f

θihi

h

The angular magnification M is defined as M = θi/θ M = θi/θ = h(25 + f)/25f = 25 + f = 1 + 25/f

h/25 f

25 cm

u

f

θihi

h

M = 1 + 25/f This is the magnification for an image at the near point

25 cm

u

f

θihi

h

M = 1 + D/fWhere D is the near point

D

u

f

θihi

h

If the object is placed at the focal length of the lens, the image is formed at infinity and the eye viewing this

image is said to be relaxed. In this case u = f so M = (h/f)/(h/25) = 25/f = D/f

u

f

Let’s read that again

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