A student is reading a lecture written on a blackboard. The lenses in her eyes have a refractive power of 59.82 diopters, and the lens-to-retina distance is 1.686 cm. (a) How far is the blackboard from her eyes? (b) If the writing on the blackboard is 4.81 cm high, what is the height of the image on her retina?

Well, the student must be quite the eye-con if she can read a lecture on a blackboard! But let's get to the question at hand. To find the distance to the blackboard, we can use the lens formula:

1/f = 1/v - 1/u

Where f is the focal length, v is the image distance, and u is the object distance. In this case, the refractive power of the lenses is equivalent to the focal length (f = 59.82 diopters = 1.68 cm^-1).

Now, we need to find the object distance, which is the distance between the student's eyes and the blackboard. Plugging in the given values, we get:

1/f = 1/v - 1/u
1/1.68 = 1/v - 1/u

Simplifying this equation, we get:

u - v = u*v/f
u*v = uf + vf
uv - vf = uf
v = u*f / (u - f)

Plugging in the given values, we get:

v = (1.686 cm) * (1.68 cm^-1) / (1.686 cm - 1.68 cm^-1)
v ≈ 1.68 cm

So, the distance to the blackboard is approximately 1.68 cm.

Now, let's move on to part (b) of the question. We're asked to find the height of the image on the retina. To do that, we'll use the magnification formula:

magnification = height of the image / height of the object = v / u

Plugging in the given values, we get:

magnification = (1.68 cm) / (1.686 cm)
magnification ≈ 0.997 cm

Therefore, the height of the image on her retina is approximately 0.997 cm.

To determine the answers to both questions, we can use the thin lens equation.

(a) The thin lens equation is given by:

1/f = 1/u + 1/v

Where:
f = refractive power of the lens = 59.82 diopters (D)
u = object distance (distance from the blackboard to the lens)
v = image distance (distance from the lens to the retina)

Since the student is reading a lecture on a blackboard, the object distance (u) is equal to the distance between her eyes and the blackboard.

Using the thin lens equation, we can rearrange it to solve for u:

1/u = 1/f - 1/v

1/u = 1/59.82 - 1/1.686

1/u = 0.016699 - 0.593182

1/u = -0.576483

Now, we can find u:

u = 1/(-0.576483)

u ≈ -1.735 cm

Since the distance cannot be negative, we take the absolute value:

u ≈ 1.735 cm

Therefore, the blackboard is approximately 1.735 cm away from the student's eyes.

(b) To find the height of the image on the retina, we can use the magnification formula:

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

The magnification is given by:

m = -v/u

Therefore:

m = -v/1.735

To find the image height (h'):

h' = magnification (m) * object height (h)

h' = m * 4.81 cm

Now, substitute the value of the magnification (m):

h' = (-v/1.735) * 4.81

Simplifying:

h' = -4.81v/1.735

Therefore, the height of the image on the retina is given by:

h' ≈ -2.772v

To find the value of v, we can rearrange the thin lens equation:

1/v = 1/f - 1/u

1/v = 0.016699 - 1/1.735

1/v ≈ 0.016699 - 0.576483

1/v ≈ -0.559784

Now, we can find v:

v = 1/(-0.559784)

v ≈ -1.787 cm

Since the distance cannot be negative, we take the absolute value:

v ≈ 1.787 cm

Finally, substitute this value into the equation for h':

h' ≈ -2.772 * 1.787

h' ≈ -4.956 cm

Since heights cannot be negative, we take the absolute value:

h' ≈ 4.956 cm

Therefore, the height of the image on the retina is approximately 4.956 cm.

To find the distance of the blackboard from the student's eyes, we can use the lens formula:

1/f = 1/v - 1/u,

where f is the focal length of the lens, v is the distance of the image from the lens, and u is the distance of the object from the lens.

In this case, the lenses in the student's eyes have a refractive power of 59.82 diopters. The formula for converting refractive power to focal length is:

focal length = 1 / refractive power.

So, the focal length of the lenses in the student's eyes would be:

f = 1 / 59.82.

The lens-to-retina distance is given as 1.686 cm, which can be considered as the distance between the lens and the image formed on the retina.

Now, we can rearrange the lens formula to solve for v:

1/f = 1/v - 1/u,

1/f + 1/u = 1/v,

v = 1 / (1/f + 1/u).

Let's substitute the values we have:

v = 1 / (1/f + 1/u),

v = 1 / (1/(1/59.82) + 1/(1.686)),

v ≈ 62.08 cm.

So, the distance of the blackboard from the student's eyes is approximately 62.08 cm.

To determine the height of the image on the retina, we can use the magnification formula:

magnification = -v/u,

where the negative sign indicates that the image is inverted.

Let's substitute the values we know:

magnification = -v/u,

magnification ≈ -62.08 cm / 1.686 cm,

magnification ≈ -36.82.

The negative magnification indicates that the image is inverted.

Now, we can find the height of the image on the retina by multiplying the magnification by the height of the object:

height of the image = magnification * height of the object,

height of the image ≈ -36.82 * 4.81 cm,

height of the image ≈ -177.33 cm.

Since the image is inverted, the negative sign indicates its orientation. Therefore, the height of the image on the retina is approximately 177.33 cm.

59.82 diopters refractive power means that 1/f = 59.82 m-1 = 0.5982 cm^-1

Her eye's focal length is thus f = 1.67 cm

Let Do be the object (blackboard) distance.

(a) 1/1.686 + 1/Do = 0.5982
1/Do = 5.07*10^-3
Do = 196 cm

(b) image height on retina:
4.81 cm*(1.686/196) = 0.041 cm