A near-sighted person might correct his vision by wearing diverging lenses with focal length f = 60 cm. When wearing his glasses, he looks not at actual objects but at the virtual images of those objects formed by his glasses. Suppose he looks at a 10-cm-long pencil held vertically 1.5 m from his glasses. WHAT IS THE HEIGHT?

Why did the near-sighted person bring a pencil to his glasses? Because he wanted to see things "write"! Now, let's solve this riddle.

If the near-sighted person's glasses have a focal length of 60 cm, they are using diverging lenses. These lenses produce virtual images rather than real images. When the person looks through the glasses, the virtual image of the pencil is formed.

To find the height of the virtual image, we can use the lens formula:

1/f = 1/v - 1/u

Where:
f is the focal length of the lens (60 cm)
v is the distance of the virtual image from the lens
u is the distance of the object from the lens

Given that the pencil is held 1.5 m (150 cm) from the glasses, and the pencil has a height of 10 cm, we need to find the height of the virtual image formed.

Let's substitute the values into the lens formula:

1/60 = 1/v - 1/150

To simplify, let's find a common denominator:

1/60 = (150 - v) / (150v)

Now, let's rearrange the equation:

150v/60 = 150 - v

Simplifying further:

5v/2 = 150 - v

Now, let's solve for v:

5v + 2v = 2(150)

7v = 300

v = 43 cm

So, the height of the virtual image formed by the glasses will be 43 cm.

To find the height of the virtual image, we can use the magnification formula:

Magnification (m) = -v/u

Where:
- m is the magnification
- v is the image distance
- u is the object distance

First, we need to find the image distance (v). Given that the person is near-sighted and wearing diverging lenses, the focal length (f) of the lenses is -60 cm. The negative sign indicates that the lenses are diverging.

Using the lens formula:

1/f = 1/v - 1/u

We can rearrange the formula to solve for v:

1/v = 1/f - 1/u

Substituting the given values:

1/v = 1/(-60 cm) - 1/(150 cm)

Simplifying:

1/v = -1/60 cm - 1/150 cm
= -5/300 cm - 2/300 cm
= -7/300 cm

Now, find the reciprocal:

v = -300 cm / 7

Using the magnification formula, we can find the magnification (m):

m = -v/u

Given that the object distance (u) is 1.5 m (or 150 cm), we have:

m = -(-300 cm / 7) / 150 cm
= 300 cm / (7 * 150 cm)
= 300 / 1050
= 2/7

Therefore, the magnification is 2/7.

To find the height of the virtual image, we can use the formula:

Height of image = Magnification * Height of object

Given that the height of the object (pencil) is 10 cm, we can calculate the height of the virtual image:

Height of image = (2/7) * 10 cm
= 20/7 cm
≈ 2.86 cm

Therefore, the height of the virtual image is approximately 2.86 cm.

To find the height of the virtual image formed by the glasses, we can use the lens formula:

1/f = 1/v - 1/u

Where:
f = focal length of the lens
v = image distance from the lens
u = object distance from the lens

Given:
f = 60 cm = 0.6 m (converted to meters)
u = 1.5 m
v = ?

To find v, we can rearrange the lens formula:

1/v = 1/f + 1/u

Calculating:
1/v = 1/0.6 + 1/1.5
1/v = 1.67 + 0.67
1/v = 2.34

Now, we can find the value of v by taking the reciprocal of both sides:

v = 1 / (1/v)
v = 1 / 2.34
v ≈ 0.43 m (rounded to two decimal places)

Now, to find the height of the virtual image, we can use the magnification formula:

magnification (m) = height of image (h') / height of object (h)

Given:
h = 10 cm = 0.1 m (converted to meters)
h' = ?

Using the magnification formula:

m = h' / h

Rearranging to solve for h':

h' = m * h

The magnification of a diverging lens is given by:

m = -v / u

Plugging in the values:

m = -(0.43 m) / (1.5 m)
m ≈ -0.29

Now, calculating the height of the virtual image:

h' = (-0.29) * (0.1 m)
h' ≈ -0.029 m

Since we are considering the height of the virtual image, we take the absolute value:

h' ≈ 0.029 m

Therefore, the height of the virtual image formed by the glasses is approximately 0.029 meters.

1/do +1/di =1/f

f= - 0.6 m (for diverging lens), do= 1.5 m
1/di=-1/do+1/f=-(1/1.5+1/0.6)=2.33
di= - 0.43 m (‘-‘ as the image is in front of the lens)
Magnification M= - di/do=0.43/1.5=0.287
h=M•H=0.287•10=2.87 cm